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2 years ago

Please solve due soon

Mathematics

2 answers:

andrezito [222]2 years ago

6 0

Answer:

Step-by-step explanation:

5+5(0)

melisa1 [442]2 years ago

4 0

The correct answer would be 5 because if x=0 then 5 0 times would just be zero

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An engineer scale model shows a building that is 3 inches tall. If the scale is 1 inch = 600 feet, how tall is the actual buildi

Sever21 [200]

The actual height of building is 1800 feet.

Step-by-step explanation:

Given,

Height of building in model = 3 inches

Scale used by engineer;

1 inch = 600 feet

Therefore;

Actual height of building = Height in model * Scale used

Actual height of building = 3 * 600

Actual height of building = 1800 feet

The actual height of building is 1800 feet.

8 0

2 years ago

The _____ of a polynomial is another name for the x- intercepts .

podryga [215]

The answer is root. Hope this helps!

6 0

2 years ago

Read 2 more answers

If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

Ivan

Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

To Prove :-

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

8 0

2 years ago

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions

ryzh [129]

Answer:

The third option is correct.

The domain of the function is given by

$Domain = 0 \leq t \leq 40$

The range of the function is given by

$Range = 0 \leq V(t) \leq 200$

Step-by-step explanation:

Sara bought a cell phone for $200 and its value has decreased at rate of $5 per month.

V(t) is the value of the phone and t is the number of months.

The domain of the function is the possible values of the number of months t.

The domain of the function is given by

$Domain = 0 \leq t \leq 40$

The range of the function is the values of V(t) that we get after substituting the possible values of the number of months t.

The range of the function is given by

$Range = 0 \leq V(t) \leq 200$

When the number of months is t = 0 then the value of the function is maximum V(t) = 200

When the number of months is t = 40 then the value of the function is minimum V(t) = 0

Therefore, the third option is correct.

6 0

3 years ago

Someone please help with this geometry question!!!

ahrayia [7]

What is the number down their?

6 0

2 years ago