ASAP help me with this

Answer:

22

Step-by-step explanation:

The slope of the curve described by the equation at the given point (0,1) as in the task content is; 4.

According to the task content, it follows that the slope of the curve can be determined by means of implicit differentiation as follows;

y⁵+ x³ = y² + 12x

5y⁴(dy/dx) -2y(dy/dx) = 12 - 3x²

(dy/dx) = (12 -3x²)/(5y⁴-2y)

Hence, since the slope corresponds at the point given; (0, 1); we have;

(dy/dx) = (12 -3(0)²)/(5(1)⁴-2(1))

dy/dx = 12/3 = 4.

Hence, slope, m = 4.

Consequent to the mathematical computation above, it can then be concluded that the slope of the curve, the line tangent to the curve at the given point is; 4.

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The two positive numbers whose difference is 9 and whose product is 2950 are 50 and 59

As a general rule, it should be noted that positive numbers are numbers that have their value greater than 0

So, we start by representing the two positive numbers with x and y.

So, we have the following equations

x - y = 9

xy = 2950

Make x the subject in the first equation x - y = 9

x = y + 9

Substitute y + 9 for x in the second equation

(y + 9) * y = 2950

Expand the equation

y^2 + 9y - 2950 = 0

Using a graphing tool, we  have the solution of the above equation to be

y = 50

Recall that:

x = 9 + y

So, we have:

x = 9 + 50

Evaluate

x = 59

Hence, the two positive numbers whose difference is 9 and whose product is 2950 are 50 and 59

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Answer:

[tex]2\frac{2}{3}[/tex]

Step-by-step explanation:

1) Convert [tex]4\frac{1}{3}[/tex] to improper fraction. Use this rule: [tex]a\frac{b}{c} =\frac{ac+b}{c}[/tex].

[tex]\frac{4\times3+1}{3} -1\frac{2}{3}[/tex]

2) Simplify 4 * 3 to 12.

[tex]\frac{12+1}{3}[/tex]

3) Simplify 12 + 1 to 13.

[tex]\frac{13}{3} -1\frac{2}{3}[/tex]

4) Convert [tex]1\frac{2}{3}[/tex]   to improper fraction. Use this rule: [tex]a\frac{b}{c} =\frac{ac+b}{c}[/tex].

[tex]\frac{13}{3} -\frac{1\times3+2}{3}[/tex]

5) Simplify 1 * 3 to 3.

[tex]\frac{13}{3} -\frac{3+2}{3}[/tex]

6) Simplify 3 + 2 to 5.

[tex]\frac{13}{3} -\frac{5}{3}[/tex]

7) Join the denominators.

[tex]\frac{13-5}{3}[/tex]

8) Simplify.

[tex]\frac{8}{3}[/tex]

9) Convert to mixed fraction.

[tex]2\frac{2}{3}[/tex]

(Decimal Form: 2.666667)

Thank you,
Eddie

The value of the given expression is approximate equal to 1.021 × 10⁴ OR 10207.8

From the question, we are to determine the value of the given expression

The given expression is

10 to the 4th times the square root of 1.042

That is,

10⁴ × √1.042

The expression can be evaluated as shown below

10⁴ × √1.042

= 10⁴ × 1.02078

= 1.02078 × 10⁴

≈ 1.021 × 10⁴

OR

= 1.02078 × 10⁴ = 10207.8

Hence, the value of the given expression is approximate equal to 1.021 × 10⁴ OR 10207.8

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[tex]x^3[/tex] is strictly increasing on [0, 5], so

[tex]\max\{x^3 \mid 0\le x\le5\} = 5^3 = 125[/tex]

and

[tex]\min\{x^3 \mid 0 \le x\le5\} = 0^3 = 0[/tex]

so the integral is bounded between

[tex]\displaystyle \boxed{0} \le \int_0^5x^3\,dx \le \boxed{125}[/tex]

Answer: no real solution

Thus, the function has no x- intercept

Step-by-step explanation:

The statement that "σ(n) < 2n holds true for all n of the form n = p²" has been proved.

Let p be any prime number, and let σ(n) be the sum of all positive divisors of the integer n.

As p is a prime number, and 2 is the smallest prime number, so, p[tex]\geq[/tex]2

So, the positive divisors of the integer n are: 1,p,p².

As σ(n) represents the sum of all positive divisors of the integer n.

σ(n)=1+p+p²

In order to prove that σ(n) < 2n,for all n of the form n = p².

1+p+p²<2p²

p²-p-1>0

It is know that, p[tex]\geq[/tex]2.

So, p²-p-1[tex]\geq[/tex]1

Thus, σ(n) < 2n holds true for all n of the form n = p².

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Solving a system of equations we will see that Hari has 800 and Ram has 600.

Let's define the variables:

We know that:

(2/3)*R = (1/2)*H

We also know that in total they have 1400, then:

R + H = 1400.

So we have the system of equations:

(2/3)*R = (1/2)*H

R + H = 1400.

In the first equation we can isolate R.

R = (3/2)*(1/2)*H = (3/4)*H

Now we can replace that in the other equation:

(3/4)*H + H  =1400

H*(7/4) = 1400

H = (4/7)*1400 = 800

So Hari has 800, and:

R + H = 1400

R = 1400 - H = 1400 - 800 = 600

Hari has 800 and Ram has 600.

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Answer: B: (1-5x²)³, all real numbers

Step-by-step explanation:

The notation [tex]f\circ g(x)[/tex] is called function composition, which is where you pass one function in as the value of another function. In other words, [tex]\( f\circ g(x)=f(g(x)) \)[/tex].

Since we have the values for f(x) and g(x), let's plug them in.

[tex]f(g(x))\\f(1-5x^2)\\(1-5x^2)^3[/tex]

Hence, the best answer choice is B, as all real numbers would work and it is the cube of 1 - 5x².

Answer:

[tex]u = x^{7} +8[/tex]     [tex]du = 7x^{6} dx[/tex]   result is [tex]\frac{1}{7} \sqrt[4]{u}[/tex]

Step-by-step explanation:

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

For two lines to be parallel, there should be angles that follow some specific properties that is usually observed with parallel lines.

We can clearly see that :

[tex] \qquad❖ \: \sf \: \angle7 \cong \angle16[/tex]

( by Alternate interior angle pair )

[tex] \qquad \large \sf {Conclusion} : [/tex]

Lines l and m are parallel to each other.

July Aug. Sept. Oct. Nov. Dec. Receipts $500 $550 $700 $850 $795 $715 Expenses $490 $550 $600 $795 $ $650 Net Cash Flow $ $0 $ $55 $45 $ Cumulative Balance $10 $ $110 $ $210 $275 the missing values are 750, 10, 100, 65, 10, 165

This is further explained below.

Generally, The term "cumulative balance" refers to the total amount of money left over at the end of a fiscal year after all surplus amounts have been subtracted from deficit amounts. If there is a negative amount in the Cumulative Balance at the conclusion of a fiscal year, then that balance will be carried forward and used as the opening balance for the next fiscal year.

The term "cumulative account" refers to the total amount of an employee's account under a defined contribution plan (for an unaggregated plan) or the total amount of an employee's account under all defined contribution plans included in an Aggregation Group (for aggregated plans), both of which are determined as of the most recent plan valuation date within the most recent 12-month period that ends on the...

In conclusion, for the following data  July Aug. Sept. Oct. Nov. Dec. Receipts $500 $550 $700 $850 $795 $715 Expenses $490 $550 $600 $795 $ $650 Net Cash Flow $ $0 $ $55 $45 $ Cumulative Balance $10 $ $110 $ $210 $275 the missing values are 750, 10, 100, 65, 10, 165

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Answer: C, E

Step-by-step explanation:

These are polynomials by the definition of a polynomial.

Answer:

SR = 5

Step-by-step explanation:

We know that the two lines are equal to each other and thus TS + SR = 13

So, we can simply set the two equations of the first line equal to 13:

x + 2 + 2x - 7 = 13

3x - 5 = 13

3x = 18

x = 6

Now we plug in 6 for x in the SR equation:

2 * 6 = 12 - 7 = 5

Answer:[tex]\Large\boxed{(2+\sqrt{2},~2-\sqrt{x} )~~and~~ (2-\sqrt{2},~2+\sqrt{x} )}[/tex]

Step-by-step explanation:

Given the system of equations

[tex]1)~xy=2[/tex]

[tex]2)~x+y=4[/tex]

Divide x on both sides of the 1) equation

[tex]xy=2[/tex]

[tex]xy\div x=2\div x[/tex]

[tex]y=\dfrac{2}{x}[/tex]

Current system

[tex]1)~y=\dfrac{2}{x}[/tex]

[tex]2)~x+y=4[/tex]

Substitute the 1) equation into the 2) equation

[tex]x+(\dfrac{2}{x} )=4[/tex]

Multiply x on both sides

[tex]x\times x+\dfrac{2}{x}\times x=4\times x[/tex]

[tex]x^2+2=4x[/tex]

Subtract 4x on both sides

[tex]x^2+2-4x=4x-4x[/tex]

[tex]x^2-4x+2=0[/tex]

Use the quadratic formula to solve for the x value

[tex]x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)} }{2(1)}[/tex]

[tex]x=2\pm\sqrt{2}[/tex]

Substitute the x value into one of the equations to find the y value

[tex]xy=2[/tex]

[tex](2+\sqrt{2} )y=2[/tex]

[tex]y=2-\sqrt{2}[/tex]

[tex]OR[/tex]

[tex]xy=2[/tex]

[tex](2-\sqrt{2} )y=2[/tex]

[tex]y=2+\sqrt{2}[/tex]

Therefore, the points of intersection are

[tex]\Large\boxed{(2+\sqrt{2},~2-\sqrt{x} )~~and~~ (2-\sqrt{2},~2+\sqrt{x} )}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

(2 - [tex]\sqrt{2}[/tex] , 2 + [tex]\sqrt{2}[/tex] ) and (2 + [tex]\sqrt{2}[/tex], 2 - [tex]\sqrt{2}[/tex] )

Step-by-step explanation:

xy = 2 → (1)

x + y = 4 ( subtract x from both sides )

y = 4 - x → (2)

substitute y = 4 - x into (1)

x(4 - x) = 2

4x - x² = 2 ( multiply through by - 1 )

x² - 4x = - 2

using the method of completing the square

add ( half the coefficient of the x- term)² to both sides

x² + 2(- 2)x + 4 = - 2 + 4

(x - 2)² = 2 ( take square root of both sides )

x - 2 = ± [tex]\sqrt{2}[/tex] ( add 2 to both sides )

x = 2 ± [tex]\sqrt{2}[/tex] , that is

x = 2 - [tex]\sqrt{2}[/tex] , x = 2 + [tex]\sqrt{2}[/tex]

substitute these values of x into (2) for corresponding values of y

x = 2 - [tex]\sqrt{2}[/tex] , then

y = 4 - (2 - [tex]\sqrt{2}[/tex])

  = 4 - 2 + [tex]\sqrt{2}[/tex]

  = 2 + [tex]\sqrt{2}[/tex] ⇒ (2 - [tex]\sqrt{2}[/tex] , 2 + [tex]\sqrt{2}[/tex] ) ← 1 point of intersection

x = 2 + [tex]\sqrt{2}[/tex] , then

y = 4 - (2 + [tex]\sqrt{2}[/tex] )

   = 4 - 2 - [tex]\sqrt{2}[/tex]

   = 2 - [tex]\sqrt{2}[/tex] ⇒ (2 + [tex]\sqrt{2}[/tex] , 2 - [tex]\sqrt{2}[/tex] ) ← 2nd point of intersection

Answer: −2(