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3 years ago

Need more math help please

Mathematics

1 answer:

Lapatulllka [165]3 years ago

4 0

The answer for this problem is D

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Is the answer one ?<br> Equation-<br> 4+x+2-2x=5

allochka39001 [22]

Combine like terms on the left side of the equation. Add 4 and 2 and subtract x and 2x. After combining like terms you should get: 6 - x = 5.

Subtract 6 from both sides to isolate x. Now you have: -x = -1. Make x into a positive by dividing both sides by -1. You are correct - the answer is x = 1.

6 0

3 years ago

WILL GIVE BRAINILY 5 STARS AND THANKS FOR CORRECT ANSWER ITS PRETTY EASY If it is 3:00 p.m. and you move the minute hand of the

Gwar [14]

Answer:

3:45 pm

Step-by-step explanation:

Every 90 degree = 15 minutes

270 degrees = 15 x 3 = 45 minutes

3:00 + 0:45 = 3:45 pm

Hope this helps!

4 0

3 years ago

Read 2 more answers

F(x)=-3x^2+9x-8 Find f(5)

Oxana [17]

Answer:

-38

Step-by-step explanation:

You replace the x with 5.

turns it into -3 (5)^2+9(5)-8

-3(25)+45-8

-75+45=-30-8=-38

4 0

4 years ago

Read 2 more answers

You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$

Thus maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

6 0

4 years ago

Freeeee points

Semenov [28]

Answer:

The answer is 8

3 0

3 years ago