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slavikrds [6]
3 years ago
10

A pen costs twice as much as a pencil. The total cost of 1 pen and 1 pencil is $2.10. If p represents the cost of 1 pencil, whic

h equation could you use to find p
A. 2p+p=10
B. p/3=2.10
C. 2p=2.10
D. p+p=2.10
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

A:2p+p=10

Step-by-step explanation:

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Jalyn collected 24 stones. She put them in 4 equal piles. How many stones are in each pile?
ZanzabumX [31]

Answer:

6

Step-by-step explanation:

This is basically 24 divided by 4 since he/she is dividing 24 into 4 equal groups.

24/4=6

3 0
3 years ago
Read 2 more answers
How to find all real solutions
AlladinOne [14]
\bf \sqrt{\sqrt{x-5}+x}=5\leftarrow \textit{squaring both sides}
\\\\
\sqrt{x-5}+x=25\implies \sqrt{x-5}=25-x\leftarrow \textit{squaring both sides}
\\\\
x-5=(25-x)^2\implies x-5=625-50x+x^2
\\\\
0=x^2-51x+630\implies 0=(x-30)(x-21)

and surely you'd know what the roots are

7 0
3 years ago
Find the volume of the wedge-shaped region contained in the cylinder x2 + y2 = 49, bounded above by the plane z = x and below by
fiasKO [112]
\displaystyle\iiint_R\mathrm dV=\int_{y=-7}^{y=7}\int_{x=-\sqrt{49-y^2}}^{x=0}\int_{z=x}^{z=0}\mathrm dz\,\mathrm dx\,\mathrm dy

Converting to cylindrical coordinates, the integral is equivalent to

\displaystyle\iiint_R\mathrm dV=\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}\int_{z=r\cos\theta}^{z=0}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}-r^2\cos\theta\,\mathrm dr\,\mathrm d\theta
=-\displaystyle\left(\int_{\theta=\pi/2}^{3\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{r=0}^{r=7}r^2\,\mathrm dr\right)
=\dfrac{2\times7^3}3=\dfrac{686}3
4 0
3 years ago
Can someone please help me??
aivan3 [116]

Answer:

-16 is the answer after evaluating

Step-by-step explanation:

{b }^{2}  - 7b - 6 \\ 5 { }^{2}  - 7 \times 5 -6 \\ 25 - 35 - 6 \\ 25 - 41 \\  - 16

4 0
2 years ago
Read 2 more answers
Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than
Fittoniya [83]
Inequation 1: 

\frac{2}{3}x-3 \geq y

to plot the pairs (x, y) for which the inequation holds, draw the line y=\frac{2}{3}x-3

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2: 

y+ \frac{2}{3}x\ \textless \ 4

y\ \textless \ - \frac{2}{3} x+ 4



draw the lines 

i)  y=\frac{2}{3}x-3          use points (0, -3),  (3, -1)

ii)y=- \frac{2}{3} x+ 4       use points ( 0, 4),   (3, 2)


let's use the point P(3, 3) to see what region of the lines need to be coloured:

\frac{2}{3}x-3 \geq y  ; 
\frac{2}{3}(3)-3 \geq 3
2-3 \geq 3, not true so we color the region not containing this point


y+ \frac{2}{3}x\ \textless \ 4
(3)+ \frac{2}{3}(3)\ \textless \ 4
3+ 5\ \textless \ 4 not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.



4 0
3 years ago
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