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Ghella [55]
3 years ago
15

Can I get some help plz.........

Mathematics
2 answers:
viva [34]3 years ago
7 0

Answer:

3/4 cm

Step-by-step explanation:

bearhunter [10]3 years ago
3 0

Answer:

108

Step-by-step explanation:

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A) not similar
inn [45]

Answer:

A) not similar

Step-by-step explanation:

for the triangles to be similar the side lengths must be proportional.

For example:

99/35 should be equal to 121/44 but when we do cross multiplying we see they are not equal therefore the the triangles are not similar.

5 0
2 years ago
Which system has no solution?<br> Check all that appy.
Ainat [17]
The first equation has no solution



The picture shows all the work I did, the reason I took a pic is because it is a little hard to explain by text, but I hope this helped you!

7 0
3 years ago
Given that 5W = 2P + 3R find the value of P when W = 4 and R = −4
Olenka [21]

Answer:

P = 16

Step-by-step explanation:

Given

5W = 2P + 3R ← substitute W = 4 and R = - 4 into the equation

5(4) = 2P + 3(- 4), that is

20 = 2P - 12 ( add 12 to both sides )

32 = 2P ( divide both sides by 2 )

16 = P

5 0
3 years ago
Ayita made 5 1/2 cups of trail mix. She want to divide the trail mix into 3/4 cup servings.How many servings will she have?
lara [203]

Answer:

7

Step-by-step explanation:


3 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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