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Allisa [31]
2 years ago
9

5c+4 = 2 (c – 5) What is the solution?

Mathematics
2 answers:
dybincka [34]2 years ago
8 0

Answer:

0

Step-by-step explanation:

if I'm correct it's 0 because when you do all of the math that's what it came out to

lukranit [14]2 years ago
3 0
C=-14/3
if that helps :)
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3. If each person paid an equal amount, who would save the most money? Explain your reasoning using at least two complete senten
Rzqust [24]

There ar 4persons so each will pay

\\ \sf\longmapsto \dfrac{83.90}{4}=20.975

Only two persons have spent more money than everyone average i.e mia and Jasmin.

Now

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Mia left more money so she saved most money.

7 0
2 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
2 years ago
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valentinak56 [21]

Answer:

Step-by-step explanation:

Is there more to this?

3 0
2 years ago
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Natali5045456 [20]
<span>f(-10)=12,      x = -10, y = 12

f(16)=-1,        x = 16, y = -1.

so, we have two points, let's check with that,

</span>\bf \begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ -10 &,& 12~) &#10;%  (c,d)&#10;&&(~ 16 &,& -1~)&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope =  m\implies &#10;\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-12}{16-(-10)}\implies \cfrac{-1-12}{16+10}&#10;\\\\\\&#10;\cfrac{-13}{26}\implies -\cfrac{1}{2}
<span>
</span>\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)} y-12=-\cfrac{1}{2}[x-(-10)]&#10;\\\\\\&#10;y-12=-\cfrac{1}{2}(x+10)\implies y-12=-\cfrac{1}{2}x-5\implies y=-\cfrac{1}{2}x+7<span>
</span>
3 0
3 years ago
Please help 1. Find the rectangular coordinates of (7, 150°).
cupoosta [38]

Answer:

Step-by-step explanation:

x=r cos \theta=7 cos150=7cos(180-30)=-7cos ~30=-\frac{7\sqrt{3} }{2} \\y=r~sin~\theta=7~sin ~150=7~sin~(180-30)=7~sin~30=\frac{7}{2} \\so ~coordinates ~are~(-\frac{7\sqrt{3}}{2} ,\frac{7}{2} )\\C

8 0
3 years ago
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