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stiv31 [10]
1 year ago
14

Does someone mind helping me with this problem? Thank you!

Mathematics
2 answers:
Ahat [919]1 year ago
5 0

Answer:

1.4 my guyyyyyyyyyyyyyyy

inessss [21]1 year ago
4 0

Answer:

0.708

Step-by-step explanation:

Change the divisor 2.1 to a whole number by moving the decimal point 1 places to the right. Then move the decimal point in the dividend the same, 1 places to the right.

We then have the equations:

14.88 ÷ 21 = 0.708

and therefore:

1.488 ÷ 2.1 = 0.708

Both calculated to 3 decimal places.

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Bought computer for $500. Marked down $50. Plus a 25% discount off original price. What is the original price?
andriy [413]
So 500+50=550

550=25-100=75%

550=75%
multiply both sdies yb 4
2200=300%
divide by 3
733.333=100%

the origonal price=$733.33
8 0
3 years ago
Read 2 more answers
What is the scale factor of ABC to DEF?
pochemuha

Answer:

A. 1

Step-by-step explanation:

<em>Scale Factor:</em>

<em>The scale factor is the number you multiply the length of a side of the first triangle to get the length of the corresponding side of the other triangle.</em>

Every side that measures 5 in triangle ABC has a corresponding side in triangle DEF that also measures 5. In this case 5 * 1 = 5, so the scale factor is 1.

Answer: A. 1

5 0
3 years ago
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Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your
BaLLatris [955]

f(x,y)=\dfrac{y^3}x

a. The gradient is

\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath

\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}

b. The gradient at point P(1, 2) is

\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}

c. The derivative of f at P in the direction of \vec u is

D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}

It looks like

\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath

so that

\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2

Then

D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}

\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}

7 0
3 years ago
I need help with this question?
Delicious77 [7]
If I remember right the answer is c
5 0
3 years ago
David bought a poster for an art project. The poster is 2.7 ft wide and 3.9 ft tall. What is the area of the poster? Enter your
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10.53 ft is the Area. :D:D
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