give the answers to the remaining boxes left blank, you can give the answers all in one sentence starting from the first blank box all the way to the last

The psi that the TPMS would trigger a warning for this car is = 23.36 psi

The target tire pressure of the car is = 32 psi (pounds per square inch.)

The Tire pressure monitoring systems (TPMS) warns the car below 27% of 32psi

That is , 27/100 × 32

= 864/100

= 8.64psi

Therefore, 32 - 8.64 = 23.36. When the car is below 23.36psi, TPMS would trigger a warning for this car.

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Complete question:

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 27% below the target pressure. Suppose the target tire pressure of a certain car is 32 psi (pounds per square inch.)

At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.) When the tire pressure is above or below?

Answer:

35

Step-by-step explanation:

first, you look at 7x, from the previous equation, you know that x=3, so you take 7x3=21 then you evaluate -3y. as you did with x on the last one you will look at the equation for y and see that it's -4. A negative times a negative is a positive, so -4x(-3)= 12. Then you add them all together, since 12 is a positive, the equation would now look like 21+12+2. After adding all three numbers together, you get 12.

Answer: 10,204 feet

Step-by-step explanation: i would assume you would add the two together, seeing as if the mountain is 10,093 above sea level and the valley is 111 below, the difference in elevation is also the distance between each other.

The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].

In this problem we have a function in implicit form, that is, an expression of the form: f(x, y, c) = 0, where c is a constant. Then, we should apply implicit differentiation twice to determine the second derivative of the function:

Original expression

x · y - 1 = 2 · x + y²

First derivative

y + x · y' = 2 + 2 · y · y'

(x - 2 · y) · y' = 2 - y

y' = (2 - y) / (x - 2 · y)

Second derivative

y' + y' + x · y'' = 2 · (y')² + 2 · y · y''

2 · y' - 2 · (y')² = (2 · y - x) · y''

y'' = 2 · [y' - (y')²] / (2 · y - x)

y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]]

The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].

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Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

Triangle is a polygon that has three sides and three angles. Types of triangles are isosceles, equilateral and scalene triangle.

Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent to each other. Also, their corresponding angles are congruent.

Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

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Answer:

  x = 23; not extraneous

Step-by-step explanation:

A solution is extraneous if it does not satisfy the original equation. Extraneous solutions can sometimes be introduced in the process of solving radical and rational function equations.

Squaring both sides of the given equation, we get ...

  √(3x +12) = 9

  3x +12 = 81 . . . . . . square both sides

  x +4 = 27 . . . . . . . divide by 3

  x = 23 . . . . . . . . . . subtract 4

There is only one solution, and it satisfies the equation:

  √(3×23 +12) = √81 = 9

The solution x = 23 is not extraneous.

Answer: x = 23; not extraneous

Step-by-step explanation:

The solution to the given expression is x = -20/11

Logarithmic function are inverse of exponential functions. Given the equation below;

log (6 x + 10) = log 1/2 x

In order to determine the solution to the given logarithmic equation, we will first have to cancel the logarithm on both sides to have

6x + 10 = 1/2x

Collect the like terms

6x - 1/2x = 0 - 10

Find the LCD

12x-x/2 = -10
11x/2 = -10

Cross multiply

11x = -2 * 10

11x = -20

Divide both sides by 11

11x/11 = -20/11

x = -20/11

Hence the solution to the given expression is x = -20/11

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Answer:

Consider the table for y= |x+1| :

x   |   y

---------

0      1

1       2

2      3

-1      2

-2     3

This would give us the parent function of y=|x| but translated up one unit. It should look like a v starting at (0, 1)

An angle is produced at the point where two or more lines meet. Thus the value of LP required in the question is approximately 14.

Two lines are said to be perpendicular when a measure of the angle between them is a right angle. While parallel lines are lines that do not meet even when extended to infinity.

From the question, let the length of LP be represented by x.

Thus, from the given question, it can be deduced that;

LM ≅ PN = 20

KP = KN - PN

    = 30 - 20

KP = 10

LP = x

Also,

<MLP is a right angle, so that;

< KLP = < KLM - <PLM

         = 126 - 90

<KLP = [tex]36^{o}[/tex]

So that applying the Pythagoras theorem to triangle KLP, we have;

Tan θ = [tex]\frac{opposite}{adjacent}[/tex]

Tan 36 = [tex]\frac{10}{x}[/tex]

x = [tex]\frac{10}{Tan 36}[/tex]

  = [tex]\frac{10}{0.7265}[/tex]

x = 13.765

Therefore the side LP ≅ 14.

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Answer:

ratio = 4

Step-by-step explanation:

According to the given table:

• the number of days with no fire incidents

  = 16

• the number of days with more than 5 fire incidents

  = 2 + 2

  = 4

Conclusion :

the ratio of the number of days with no fire incidents

to the number of days with more than 5 fire incidents is :

16 to 4 (16 : 4)

Then

The ratio = 4

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

Equation of directrix is : y = 1, so we can say that it's a parabola of form : -

[tex]\qquad \sf  \dashrightarrow \: (x - h) {}^{2} = 4a(y - k)[/tex]

and since the focus is above the directrix, it's a parabola with upward opening.

[tex]\qquad \sf  \dashrightarrow \: (x - ( - 4)) {}^{2} = 4(2)(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: (x + 4) {}^{2} = 8(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: {x}^{2} + 8x + 16 = 8y - 40[/tex]

[tex]\qquad \sf  \dashrightarrow \: 8y = {x}^{2} + 8x + 56[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = \cfrac{1}{8} {x}^{2} + x + 7[/tex]

Directrix

Focus

Focus lies in Q3 and above y=1

Then

Perpendicular distance

Find a for the equation

Now the equation is

[tex]\\ \rm\dashrightarrow 4a(y-k)=(x-h)^2[/tex]

[tex]\\ \rm\dashrightarrow 4(2)(y-5)=(x+4)^2[/tex]

[tex]\\ \rm\dashrightarrow 8(y-5)=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y-40=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+16+40[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+56[/tex]

[tex]\\ \rm\dashrightarrow y=\dfrac{x^2}{8}+x+7[/tex]

Answer:

t2=8/5

Step-by-step explanation:

using this formula

t1/s1 =t2/s2

4/5=t2/2

cross multiply

5t2=8

t2=8/5

The correct answer for the value of t₂ is [tex]1.6[/tex].

Given:

Time t₁ = 4,

Distance s₂ =2

Distance s₁ = 5.

To find value of t₂ , use the concept of proportion:

[tex]\dfrac{t_1}{s_1} = \dfrac{t_2}{s_2}[/tex]

Put value of [tex]t_1 ,s_1 ,s_2[/tex]:

[tex]\dfrac{t_2}{2} =\dfrac{4}{5}\\\\t_2 =\dfrac{8}{5}\\\\ t_2 = 1.6[/tex]

The correct value of [tex]t_2[/tex] is [tex]1.6[/tex].

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Answer:

Hello,

Step-by-step explanation:

All is in the picture.

B=(-2,8), O=(-2,-5)

b=BO=8+5=13

F_1=(-2,4)   O=(-2,5)  Focus distance=4+5=9
Horizontal half axis=√(b²-f²)=√88

the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

Event A is rolling a sum less than 6.

Let's define the possible elements in this experiment as:

(outcome of dice 1, outcome of dice 2)

The outcomes where the sum is less than 6 are:

dice 1    dice 2     sum

  1               1           2

  1               2           3

  1               3          4

  1               4          5

  2               1          3

  3              1           4

  4               1          5

  2              2          4

  3              2          5

  2              3           5

So there are 10 outcomes, then the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

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Answer:

B) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

Step-by-step explanation:

If the question said less than 6 meaning you have to find all possible solution that are 5 or lower.

However, if the problem said equal or less than 6 then you have to find all possible solution that are 6 or lower.

B option is only option that don't have sum of 6. Therefore, option B is correct.

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

The correct value of w is 20 as the width of a rectangle must be positive. A quadratic function always has 2 zeroes and in a case like this the negative one is ignored.

The arc length is given by the definite integral

[tex]\displaystyle \int_1^3 \sqrt{1 + \left(y'\right)^2} \, dx = \int_1^3 \sqrt{1+9x} \, dx[/tex]

since by the power rule for differentiation,

[tex]y = 2x^{3/2} \implies y' = \dfrac32 \cdot 2x^{3/2-1} = 3x^{1/2} \implies \left(y'\right)^2 = 9x[/tex]

To compute the integral, substitute

[tex]u = 1+9x \implies du = 9\,dx[/tex]

so that by the power rule for integration and the fundamental theorem of calculus,

[tex]\displaystyle \int_{x=1}^{x=3} \sqrt{1+9x} \, dx = \frac19 \int_{u=10}^{u=28} u^{1/2} \, du = \frac19\times\frac23 u^{1/2+1} \bigg|_{10}^{28} = \boxed{\frac2{27}\left(28^{3/2} - 10^{3/2}\right)}[/tex]

Angles is known to be opposite each other when two lines cross called vertical angles  due to the fact that they are opposite each other at a vertex.

Vertical Angles are known to be often called Vertically Opposite Angles and this is described as the scenario when two lines intersect one another, then the opposite angles, is made as a result of the intersection which is known to be called vertical angles or what we say as vertically opposite angles.

Note that A pair of vertically opposite angles are said to be often always equal to one another.

Hence, based on the scenario above, Angles is known to be opposite each other when two lines cross called vertical angles  due to the fact that they are opposite each other at a vertex.

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Answer: c) {1,245; 1,224; 1,203; 1,182; …}

Step-by-step explanation:

Concept:

For this question, we just go by eliminating each answer until we get the correct one

Given information

Common difference = -21 (decreasing sequence)

Answer Choice: a) {873, 894, 915, 936, …}

894 - 873 = 21

915 - 895 = 21

936 - 915 = 21

Since the common difference is 21, not -21

[tex]\large\boxed{FALSE}[/tex]

Answer Choice: b) {32, 20, 8, -4, …}

20 - 32 = -12

8 - 20 = -12

-4 - 8 = -12

Since the common difference is -8, not -21

[tex]\large\boxed{FALSE}[/tex]

Answer Choice: c) {1,245; 1,224; 1,203; 1,182; …}

1224 - 1245 = -21

1203 - 1224 = -21

1182 - 1203 = -21

Since the common difference is -21

[tex]\Huge\boxed{TRUE}[/tex]

Answer Choice: d) {1,563; 1,587; 1,611; 1,635; …}

1587 - 1563 = 24

1611 - 1587 = 24

1635 - 1611 = 24

Since the common difference is 24, not -21

[tex]\large\boxed{FALSE}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer

{-10, 9}

Step-by-step explanation:

The above are Simultaneous equations

These are two or more equations that share same variables.

8X + 10Y = 10--------(1)

5X + 4Y = -14-------(2)

40X + 50Y = 50 ---------(3)

40X + 32Y = - 112--------(4)

18Y = 162

[tex] \frac{18y}{18} = \frac{162}{18} \\ y = 9[/tex]

40X + 50Y = 50

40X + 50(9) = 50

40X + 450 = 50

40X = 50 - 450

40X = - 400

{-10, 9}

Usando la distribución hipergeométrica,  la probabilidad de que ambas estén numeradas con el valor 1, es: 0.0222 = 2.22%.

La fórmula es:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Los parámetros son:

Los valores de los parámetros son:

N = 10, k = 2, n = 2.

La probabilidad de que ambas estén numeradas con el valor 1, es P(X = 2), entonces:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,10,2,2) = \frac{C_{2,2}C_{8,0}}{C_{10,2}} = 0.0222[/tex]

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The solutions for the given equations are:

Writing a number or an equation as a product of its factors is said to be the factorization.

A linear equation has only one factor, a quadratic equation has 2 factors and a cubic equation has 3 factors.

1. Solving x² - 49 = 0; (quadratic equation)

⇒ x² - 7² = 0

This is in the form of a² - b². So, a² - b² = (a + b)(a - b)

⇒ (x + 7)(x - 7) =0

By the zero-product rule,

x = -7 and 7.

2. Solving 3x³ - 12x = 0

⇒ 3x(x² - 4) = 0

⇒ 3x(x² - 2²) = 0

⇒ 3x(x + 2)(x - 2) = 0

So, by the zero product rule, x = -2, 0, 2

3. Solving 12x² + 14x + 12 = 18; (quadratic equation)

⇒ 12x² + 14x + 12 - 18 = 0

⇒ 12x² + 14x - 6 = 0

⇒ 2(6x² + 7x - 3) = 0

⇒ 6x² + 9x - 2x - 3 = 0

⇒ 3x(2x + 3) - (2x + 3) = 0

⇒ (3x - 1)(2x + 3) = 0

∴ x = 1/3, -3/2

4. Solving -x³ + 22x² - 121x = 0

⇒ -x³ + 22x² - 121x = 0

⇒ -x(x² - 22x + 121) = 0

⇒ -x(x² - 11x - 11x + 121) = 0

⇒ -x(x(x - 11) - 11(x - 11)) = 0

⇒ -x(x - 11)² = 0

∴ x = 0, 11, 11

5. Solving x² - 4x = 5; (quadratic equation)

⇒ x² - 4x - 5 = 0

⇒ x² -5x + x - 5 = 0

⇒ x(x - 5) + (x - 5) = 0

⇒ (x + 1)(x - 5) =0

∴ x = -1, 5

Hence all the given equations are solved.

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Answer:

yes

Step-by-step explanation:

25x² - 40xy + 16y² can be factored as

(5x - 4y)² ← a perfect square

Rewriting the left-hand side as follows,

[tex](2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40[/tex]

Answer:  d: 1:52pm

Step-by-step explanation:  Since Beatrice finished 22 minutes earlier, we subtract 22 minutes from 2:14. 2:14 - 14 is 2:00. 22-14 is 8. 2:00 - 8 is 1:52.

The lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

The expressions are given as:

3xyz^2 and 9x^2y + 9x^2

Factorize the expressions

3xyz^2 = 3 * x * y * z * z

9x^2y + 9x^2 = 3 * 3 * x * x * (y + 1)

Multiply the common factors, without repetition

LCM = 3 * 3 * x * x * (y + 1) * z* z

Evaluate the product

LCM = 9x^2z^2(y + 1)

Hence, the lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

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Answer:

See below for proof.

Step-by-step explanation:

Given:

[tex]y=\left(x+\sqrt{1+x^2}\right)^m[/tex]

First derivative

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}[/tex]

[tex]\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}[/tex]

Second derivative

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}[/tex]

[tex]\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}[/tex]

[tex]\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}[/tex]

              [tex]= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)[/tex]

Proof

  [tex](x^2+1)y_2+xy_1-m^2y[/tex]

[tex]= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right][/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[0][/tex]

[tex]= 0[/tex]

Answer:

y= -1

Step-by-step explanation:

(2) (0) − y = 1

0 + − y = 1

(−y) + (0) = 1

−y = 1

Step 2: Divide both sides by -1

Y = −1