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boyakko [2]
3 years ago
15

Which equation shows P=2l + 2w when solved for w?

Mathematics
2 answers:
larisa86 [58]3 years ago
7 0

Answer:

Is d. W=P-2L/2

Jlenok [28]3 years ago
3 0

Answer:

the answer is A

Step-by-step explanation:

when solving for a literal equation you wanna get the letter by itself

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Is 2/9 equal to 15/64? And how do I figure it out?​
svlad2 [7]

Answer:

no

Step-by-step explanation:

you can't multiply any number by 2 to get 15

5 0
3 years ago
What number divided by -7 is equal to -15?
Ilia_Sergeevich [38]
105 is the number
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4 0
3 years ago
Find the positive value of x if 4 - 2x] = 6.
Alecsey [184]

6-4+2x=0

2+2x=0

-2(-1-x)=0

Step-by-step explanation:

not possible

4 0
3 years ago
Help me please someone
Andre45 [30]
12.
To find perimeter of triangles, use the formula P = a + b + c, where a, b, and c are the sides of the triangle.

For this triangle, you are given two sides and the perimeter and asked to find the third side. To do that, plug the given values into the formula for perimeter and solve.

P = a + b + c
10.8 = 3.5 + 2.6 + c
10.8 = 6.1 + c
4.7 = c

The measure of the third side of the triangle is 4.7 cm.

13.
The perimeter of a quadrilateral is the sum of the length of its four sides. So to find the perimeter, simply add all the sides together. As you need to find a missing side, follow the same steps as above to find the missing side.

15.3 = 2.7 + 4.7 + 3.3 + x
15.3 = 10.7 + x
4.6 = x

The length of the missing side of the quadrilateral is 4.6 cm.
7 0
4 years ago
Who can help me d e f thanks​
12345 [234]

d)

y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}

Product rule:

y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'

Chain and power rules:

y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'

Power rule:

y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)

Now simplify.

y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

e)

y = \dfrac{3bx + ac}{\sqrt{ax}}

Quotient rule:

y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}

y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}

Power rule:

y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}

Now simplify.

y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}

y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}

y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}

f)

y = \sin^2(ax+b)

Chain rule:

y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'

y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'

y' = 2a \sin(ax+b) \cos(ax+b)

We can further simplify this to

y' = a \sin(2(ax+b))

using the double angle identity for sine.

7 0
2 years ago
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