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3 years ago

5

Jenna's new horse poster has 4 right angles and 2 pairs of parallel sides. Which of the following correctly describes the shape

of the poster shown below?

2 answers:

Art [367]3 years ago

6 0

I think it’s Rectangle.

max2010maxim [7]3 years ago

3 0

Answer:

It's both a parallelogram and rectangle.

Step-by-step explanation:

You might be interested in

10(2^×) + 7(3^×) = 6^× + 70

Mkey [24]

Answer:

x = 2.81 and 2.096

Step-by-step explanation:

Given the expression

10(2^x) + 7(3^x) = 6^x + 70

This can also be expressed as;

10(2^x) + 7(3^x) = (2*3)^x + 70

10(2^x) + 7(3^x) = 2^x*3^x + 70

Let a = 2^x and b = 3^x

10a + 7b = ab + 70

10a + 7b - ab = 70

10a-ab + 7b - 70 = 0

a(10-b)+7(b-10) = 0

a(10-b)-7(10-b) = 0

a-7 = 0 and 10-b = 0

a = 7 and b = 10

Since a = 2^x

7 = 2^x

log 7 = log2^x

log7 = xlog2

x = log7/log2

x = 2.81

Similarly

10 = 3^x

log 10 = log 3^x

log 10 = xlog3

x = log 10/log 3

x = 1/0.4771

x = 2.096

Hence the values of x that satisfies the equation are 2.81 and 2.096

8 0

2 years ago

A salesperson earns a 10% commission on the selling price of each vehicle after a $2,000 dealership fee is deducted from that pr

Ipatiy [6.2K]

Answer:

Salesprice =$25000

Step-by-step explanation:

comission=$ 2300;10

salespreice=x:100

x=2300*100/10=$23000

dealership fee= $2000

salesprice= $25000

5 0

3 years ago

Which expression can be used to find the radius of a circle with a diameter of 90 inches?

rjkz [21]

The answer to this is C, 90/2

6 0

2 years ago

Read 2 more answers

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

Multiply the sum of -13/8 and 5/12 by their difference.

REY [17]

Answer:

1421/576

Step-by-step explanation:

Sum = - 13/8 + 5/12 = - 39/24 + 10/24 = - 29/24

Difference = - 13/8 - 5/12 = - 39/24 - 10/24 = - 49/24

Sum * Difference = (-29/24)*(-49/24) = 1421/576

6 0

2 years ago