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Monica [59]
3 years ago
5

What is the answer to this Question? 8 – 6 (x – 4) = 2x – 2 (4x – 5) + 22

Mathematics
1 answer:
zimovet [89]3 years ago
6 0

Answer:

Here

Step-by-step explanation:

Let's solve your equation step-by-step.

8−6(x−4)=2x−2(4x−5)+22

Step 1: Simplify both sides of the equation.

8−6(x−4)=2x−2(4x−5)+22

8+(−6)(x)+(−6)(−4)=2x+(−2)(4x)+(−2)(−5)+22(Distribute)

8+−6x+24=2x+−8x+10+22

(−6x)+(8+24)=(2x+−8x)+(10+22)(Combine Like Terms)

−6x+32=−6x+32

−6x+32=−6x+32

Step 2: Add 6x to both sides.

−6x+32+6x=−6x+32+6x

32=32

Step 3: Subtract 32 from both sides.

32−32=32−32

0=0

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What is 3x2 [-2 3 0. -4 2 6. 6 -5 6.]
densk [106]

Answer:

3 x 2 = 6

Step-by-step explanation:

By multiplying both 3 and 2, it equals to 6.

Hope this helps! :)

7 0
3 years ago
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cSuppose you are standing such that a 45-foot tree is directly between you and the sun. If you are standing 200 feet away from t
beks73 [17]

Answer:

you could stand at 5.0 ft and still be completely in the shadow of the tree

Step-by-step explanation:

From the diagram attached below;

We consider;

\overline {BC} to be the height of the tree and \overline {DE} to be the height of how tall you could be and still be completely in the shadow of the tree.

∠D = ∠B = 90°

Also;

ΔEAD = ΔBAC   (similar triangles)

Therefore, their sides will also be proportional

i.e

\dfrac{\overline {DE}}{ \overline {BC}}= \dfrac{\overline{AD}}{ \overline{AC}}

\dfrac{x}{ 45}= \dfrac{225-220}{225}

\dfrac{x}{ 45}= \dfrac{25}{225}

By cross multiply

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x = \dfrac{45 \times 25}{225}

x = \dfrac{1125}{225}

x = 5.0 ft

Therefore, you could stand at 5.0 ft and still be completely in the shadow of the tree

3 0
3 years ago
Charlie Runs out of speed of 3 yd./s. About how many miles per hour does Charlie run
grigory [225]

3 yds/sex*3600 sec/hr=10800yds/hr

10800yds/hr/1760yds/mile

=6.136363 miles/hr  


4 0
3 years ago
Please help me I will give you bring me if it’s right
Otrada [13]

Answer:x

Step-by-step explanation:

7 0
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Help with this please ..
erastova [34]
N=9
Use the distributive property
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