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3 years ago

There are twelve shirts in your closet,

Mathematics

1 answer:

Gnoma [55]3 years ago

7 0

Answer:

what is the question

Step-by-step explanation:

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Please help me out if you can :)

mojhsa [17]

Step-by-step explanation:

here,

x°+15°=5x°-41°[being vertically opposite angle]

15°+41°=5x°-x°

56°=4x°

$\frac{56 }{4x}$

thus,x=14°

<h3>hence,the value of x is 14°.</h3><h2>I hope this will help you.</h2><h2>please mark my answer as BRAINLIEST if you liked it....</h2>

8 0

3 years ago

Read 2 more answers

-2(x - 14) help me? I need help so I can get my grade up in algebra and pass my class so please help me?

sukhopar [10]

Answer:

$- 2x + 28$

Step-by-step explanation:

1) Expand by distributing terms.

$- 2x - 2 \times - 14$

2) Simplify 2 × -14 to -28.

$- 2x - ( - 28)$

3) Remove parentheses.

$- 2x + 28$

Therefor, the answer is - 2x + 28.

4 0

3 years ago

If Ill m, solve for x. (9x+2) 119° x =

Sav [38]

Answer:

x=13

Step-by-step explanation:

119-2=117

117÷9=13

9×13+2=119

3 0

3 years ago

20 points!

meriva

All circles are similar because all circle have same shape.

However Two circles are congruent , if and only if their radii are congruent.

In our question we are asked when are all circles similar given if their radii are congruent.

So we can say this statement is false because no matter the radii are congruent or not , two or more circles are always similar because of their same shape no matter what the measure of their radii is. Only if we are asked when they are congruent, we will consider the radii part.

Answer is false.

4 0

4 years ago

Read 2 more answers

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago