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Neko [114]
3 years ago
8

Please correct my errors if there are any

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

it looks good to me

Step-by-step explanation:

But i have a feeling that 8 might not be in the right place but im not sure

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It costs the quality company $.14 per mile to run its van. On Monday the van traveled 87.6 miles. How much did it cost to run th
aksik [14]
We can find this answer by multiplying .14 by 87.6 to represent 14 cents for every mile.

87.6
<u>  .14
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Since we are rounding to the nearest cent, that being the hundredths place, we know that it cost $12.26 to run the van that day. 


8 0
4 years ago
Read 2 more answers
–4(5x – 9) + 6(2x + 5)?
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-2(2(5x - 9) -3 (2x + 5))
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3 0
3 years ago
18. Physical Science A ball is dropped from a height
Alexus [3.1K]

Answer:

104.8576

Step-by-step explanation:

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5 0
3 years ago
Given an equation: y=3x-2, write an equation of a parallel line that includes the point (1,5)
Nuetrik [128]

Answer:

y = 3x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

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with slope m = 3

Parallel lines have equal slopes , then

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To find c substitute (1, 5 ) into the partial equation

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7 0
3 years ago
Not sure f either of my answers would satisfy the question in the second picture
xz_007 [3.2K]
\bf \begin{array}{llll}&#10;tan(\alpha)=-\cfrac{80}{15}\qquad II\\\\&#10;x=-15\\&#10;y=8\\&#10;r=17&#10;\end{array} \qquad \qquad &#10;\begin{array}{llll}&#10;cos(\beta)=\cfrac{5}{6}\qquad I\\\\&#10;x=5\\&#10;y=\sqrt{11}\\&#10;r=6&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})&#10;\\\\\\&#10;sin({{ \alpha}} + {{ \beta}})=\cfrac{8}{17}\cdot \cfrac{5}{6}+\cfrac{-15}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{40-15\sqrt{11}}{102}



\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})&#10;\\\\\\&#10;\cfrac{-15}{17}\cdot \cfrac{5}{6}-\cfrac{8}{17}\cdot \cfrac{\sqrt{11}}{6}\implies \cfrac{-75-8\sqrt{11}}{102}





\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}&#10;\\\\\\&#10;tan({{ \alpha}} + {{ \beta}}) = \cfrac{-\frac{8}{15}+\frac{\sqrt{11}}{5}}{1-\left( -\frac{8}{15}\cdot \frac{\sqrt{11}}{5} \right)}\implies &#10;\cfrac{\frac{-8+3\sqrt{11}}{15}}{1+\frac{8\sqrt{11}}{75}}&#10;\\\\\\&#10;tan({{ \alpha}} + {{ \beta}}) =\cfrac{\frac{-8+3\sqrt{11}}{15}}{\frac{75+8\sqrt{11}}{75}}\implies \cfrac{-8+3\sqrt{11}}{15}\cdot \cfrac{75}{75+8\sqrt{11}}&#10;\\\\\\&#10;

\bf tan({{ \alpha}} + {{ \beta}}) =\cfrac{-8+3\sqrt{11}}{1}\cdot \cfrac{3}{75+8\sqrt{11}}\implies \cfrac{15\sqrt{11}-40}{75+8\sqrt{11}}
6 0
4 years ago
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