All categories

3 years ago

1. What is AC? AC = 6 AC = 24 AC = 8 AC = 16

Mathematics

1 answer:

Alenkinab [10]3 years ago

5 0

Answer:

AC = 24

Step-by-step explanation:

If you subtract C - A, you get

16 - (-8) = 16 + 8 = 24

You might be interested in

What graph represents the equationy = 2x – 3?

gladu [14]

Answer:

linear graph

Step-by-step explanation:

because the equation of the line is straight

8 0

3 years ago

Morning can someone help me please and can you please let me know how you got the answer

Westkost [7]

Answer:

The customary system states that one pound is equivalent to sixteen ounces(oz).

Step-by-step explanation:

Since the package is one pound, it contains sixteen ounces(oz).

7 0

3 years ago

Jose asks his friends to guess the higher of two grades he received on his math tests. He gives them two hints. The difference o

denis23 [38]

Option D. 96 is the correct answer.

Explanation:

Let the higher grade be = x

Let the lower grade be = y

As given, The difference of the two grades is 16

$x-y=16$ or $x=16+y$ .... (1)

The sum of one-eighth of the higher grade and one-half of the lower grade is 52.

$\frac{1x}{8}+\frac{1y}{2}=52$ ... (2)

Putting the value of x=16+y in equation (2)

$\frac{16+y}{8}+\frac{y}{2}=52$

$\frac{16+y+4y}{8}=52$

$\frac{16+5y}{8}=52$

$16+5y=416$

$5y=400$

$y=80$

As x=16+y

x=16+80 = 96

Hence, higher grade is 96.

8 0

3 years ago

Solve for y: 5x-4y=28

Dmitriy789 [7]

Answer:

y = $\frac{5}{4} x - 7$

Step-by-step explanation:

Step 1: Add -5x to both sides.

5x − 4y + −5x = 28 + −5x

−4y = −5x + 28

Step 2: Divide both sides by -4.

$\frac{-4y}{-4} =\frac{-5x +28}{-4}$

y = $\frac{5}{4}x -7$

6 0

3 years ago

Read 2 more answers

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$

Since $L^{''}(x)$ is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

$L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}$

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

6 0

3 years ago