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aliya0001 [1]
3 years ago
6

Please answer this I REALLY REALLY REALLY NEED IT....

Mathematics
1 answer:
ziro4ka [17]3 years ago
8 0

The answers to this questions would be:

1. A-> (1.5, 1.5)                              5. A and B->  7 units  

2. B-> (-2, -1.5)                               6. C and D->  3 units  

3. C-> (-3.5, 2)                                7. B and D-> 1 unit

4. D-> (0.5, 0)                                 8. A and D-> 8 units

-------------------------------- WORD PROBLEMS------------------------------------------

9. 8 feet

10a. (-4, -1)                                              10b. (4, 1)


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Find the equation f(x) = a(x - h)2 + k for a parabola containing point (3, 6) and having (1, -2) as a vertex. What is the standa
nlexa [21]

f(x) = 2 x^{2} -4x

Step-by-step explanation:

Step 1 :

Given, f(x) = a(x - h)2 + k

Point on the parabola is (3, 6)

Vertex (h,k) = (1,-2)

Step 2:

Substituting the vertex in the equation we have,

f(x) = a(x-1)2 -2

Substituting the point (3,6) in this we have,

6 = a(3-1)2 - 2 => 6 = 4a -2

=>  4a = 8 => a = 2

Step 3 :

Substituting the value for a and the vertex in the given equation we have

f(x) = 2(x-1)2 -2 = 2(x2 - 2x + 1) -2 = 2x2 - 4x

=> f(x) = 2 x^{2} -4x  which is the standard form

8 0
3 years ago
Anyone know the answer ?
Firdavs [7]

Answer:

7

Step-by-step explanation:

10x-2=9x+5 because the angles are the same as m and n are parallel.

10x=9x+7

x=7

8 0
2 years ago
Read 2 more answers
How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title
kotegsom [21]
Let t=\sqrt y, so that t^2=y, t^4=y^2, and \mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}. Then

\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}

Now let y=\sqrt2\tan z, so that \mathrm dy=\sqrt2\sec^2z\,\mathrm dz. Then

\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C

Transform back to y to get

\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C

and again to get back a result in terms of t.

\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C
3 0
3 years ago
Factor using GCF or monomialpfactoring:<br> w3 + 3w<br> w3(w + 3)<br> W(w2 + 3)<br> 3W(1)
liraira [26]

Answer:

w(w² + 3)

Step-by-step explanation:

Given the expression

w³ + 3w

Since w is common to the terms w³ and 3w, we will factor it out from the expression as shown;

w³ + 3w

w(w²)+ 3(w)

w(w² + 3)

This gives the required expression

7 0
3 years ago
Explain how to write a addition sentence for a picture of 4 rows with 3 items in each row
rosijanka [135]
3+3+3+3 is the answe because there are 4 rows of three
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3 years ago
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