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agasfer [191]
3 years ago
8

A newsboy notices he has four times as many dimes as quarters. if his dimes and quarters total$1.95, how many of each has he?

Mathematics
1 answer:
MrRissso [65]3 years ago
4 0
D=4q
25q+10d=195
5q+2d=39

sub 4q foro d
5q+2(4q)=39
5q+8q=39
13q=39
divide 13
q=3

d=4q
d=4(3)
d=12

12 dimes
3 quarters
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Between permanent life insurance and term life insurance, which typically has the lower premium and why?
adoni [48]

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5 0
2 years ago
To which graph does the point (−1, 3) belong? y ≥ 2x + 6 y ≥ 3x − 1 y ≥ −x + 3 y ≥ −2x + 5
Anna35 [415]
Well first find the slope of the point
y=mx+b  y=-1x+3
slope = -1
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4 0
3 years ago
Read 2 more answers
Arks
luda_lava [24]

Answer:

EXPLAIN

Step-by-step explanation:

6 0
2 years ago
Find the sum of the constants a, h, and k such that
ra1l [238]

Answer:

  3

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

__

<u>Solve for h</u>

Now, we have ...

  2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

  = 2(x^2 -2hx +h^2) +k

  = 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

  -4hx = -8x

  h = (-8x)/(-4x) = 2

__

<u>Solve for k</u>

Now, we're left with ...

  2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

  k = 7 -8 = -1

__

And the sum of constants a, h, and k is ...

  a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

6 0
3 years ago
WHO CAN solve it Please !
Mariulka [41]

Answer:

a) True

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em>            </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha<em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em>            cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the  integration</em>

<em>         </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha<em></em>

<em>                                      = </em>\int\limits^\pi _0 {cos\alpha } \, dx<em></em>

<em>Now, Integrating </em>

<em>                                  </em>= ( sin\alpha )_{0} ^{\pi }<em></em>

<em>                                = sin π - sin 0</em>

<em>                               = 0-0</em>

<em>                              = 0</em>

<u><em>Final answer:-</em></u>

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

<em></em>

<em></em>

5 0
2 years ago
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