If f(x) = x2 + 7, what is the value of f(4)?

A muffin recipe calls for 2/5 tablespoons of vanilla extract for 6 muffins. Arthur is making 18 muffins. How much vanilla extrac

GuDViN [60]

2/5 tablespoons of vanilla extract ... 6 muffins
x tablespoons of vanilla extract = ? ... 18 muffins

If you would like to know how much vanilla extract does he need, you can calculate this using the following steps:

x * 6 = 18 * 2/5
x * 6 = 36/5 /6
x = 36/5 / 6
x = 6/5 = 1 1/5

Result: Arthur will need 1 1/5 tablespoons of vanilla extract for 18 muffins.

7 0

3 years ago

A department store sells logo shirts at an original price of $20. Every month that a shirt doesn’t sell, the store reduces the s

Tasya [4]

Given:
Original price = 20
reduces selling price by 25% every month it's not sold.

First markdown month:
20 * (100%-25%) = 20 * 75% = 15

Second markdown month
15 * 75% = 11.25

Macy, employee gets a 50% discount off the current price.
11.25 * 50% = 5.625
11.25 - 5.625 = 5.625 or 5.63

The pre-tax price of the shirt for Macy will be $5.63

3 0

2 years ago

Read 2 more answers

I am confused about how to do this problem and I don't think it should be hard, but I just don't know how to approach it, so if

solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) (0.875, 0.676)

d) (0, 1.235)

Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8. To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

5 0

2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago

Someone please help explain how to find x

dimaraw [331]

Answer:

x=27

Step-by-step explanation:

180-2x-40+180-3x+180-x-50+180-4x = 360

360-2x-40-3x+360-x-50-4x= 360

360-10x-90= 0

10x= 270

x= 27

5 0

3 years ago

If f(x) = x2 + 7, what is the value of f(4)?​

If f(x) = x2 + 7, what is the value of f(4)?