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AnnZ [28]
3 years ago
15

Of two numbers, one is 7 more than the other. Find the numbers if their sum is 21

Mathematics
2 answers:
BabaBlast [244]3 years ago
5 0
Hi there! The smallest number is 7 and the largest number is 14.

Let the smallest number be represented by X.
The largest number is therefore equal to X + 7.

Now we can set up and solve an equation.
x + x + 7 = 21

Collect terms
2x + 7 = 21

Subtract 7 from both sides.
2x = 14

Divide both sides by 2.
x = 14 \div 2 = 7

Hence, the smallest number is 7 and the largest number is 7 + 7 = 14.
RideAnS [48]3 years ago
4 0
It would be 14 and 7.
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Which equation represents a line that has a slope of 3/4 and passes through the point (2;1)?
omeli [17]

Answer:

B

Step-by-step explanation:

m = 3/4

x1 = 2

y1 = 1

Using the formula,

y - y1 = m(x - x1)

y - 1 = 3/4(x - 2)

y - 1 = 3x/4 - 3/2

y = 3x/4 -3/2 + 1

y = 3x/4 - 1/2 (×4)

4y = 3x - 2

Therefore, the answer is B

8 0
3 years ago
Solve each. Show all necessary work.
LenaWriter [7]

Answer:

y = 3x - 9

Step-by-step explanation:

First, find the slope using y2 - y1 / x2 - x1

36 - 21 = 15

15 - 10 = 5

15/5 = 3, so the slope (m) is 3.

Now, plug into point-slope form: y - y1 = m (x - x1)

y - 21 = 3 (x - 10)

Simplify

y - 21 = 3x - 30

y = 3x - 9

I hope this helps!!

8 0
3 years ago
Which of the following is a like radical to 3x sqrt 5
lara31 [8.8K]

Answer:

no

Step-by-step explanation:

no

4 0
3 years ago
Read 2 more answers
CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS
finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve y=x^2, with 1\le x\le2, and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6

But perhaps you'd like the surface integral treatment. Parameterize the surface by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k

with 1\le u\le2 and 0\le v\le2\pi, where the third component follows from

z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2

Take the normal vector to the surface to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}

Then the area of the surface is

\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

which reduces to the integral used in the surface-of-revolution setup.

7 0
3 years ago
Will give Brainliest!!!
professor190 [17]

Answer:

5 x T = L

Step-by-step explanation:

8 0
3 years ago
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