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Reptile [31]
2 years ago
8

Please help me out and answer this i need it ASAP <3​

Mathematics
1 answer:
QveST [7]2 years ago
6 0

Answer:

y-1=\frac{5}{6}\left(x+8\right)

I hope this is good enough:

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The longest side of an acute triangle measures 30 inches. The two remaining sides are congruent, but their length is unknown.
lakkis [162]

The smallest perimeter is 72.44 inches. The longest side of an acute triangle measures 30 inches. The two remaining sides are congruent, but their length is unknown. As much as possible, the sum of the length of the two remaining side must be greater than the other side. Given the side is 30, then the sum of the two remaining sides should be greater than 30 inches.

7 0
3 years ago
PLEASE HELP. <br> I know how to do these but I'm stuck on this one.
Nezavi [6.7K]

Steve ran 11 miles

let the distance Steve ran be x

Then distance Kevin ran is x + 4 ( 4 miles more than Steve )

x + x + 4 = 26 ( the sum of their distance is 26 )

2x + 4 = 26 ( subtract 4 from both sides )

2x = 22 ( divide both sides by 2 )

x = 11

Steve ran 11 mies and Kevin ran 11 + 4 = 15 miles

check 11 + 15 = 26


5 0
3 years ago
Help meee please <br><img src="https://tex.z-dn.net/?f=%20%20%5Csqrt%7B%3F%7D%20%7B78%7D%20%7D%20" id="TexFormula1" title=" \sq
tekilochka [14]
I think it is 8.83? I am not positive but it is worth a try
7 0
2 years ago
Increase £110 by 10%
UNO [17]

Answer:

121

Step-by-step explanation:

  1. Find 10% of 110: \frac{10}{100} × \frac{110}{1} = \frac{11}{1} = 11  
  2. Increase means add: 110 + 11 = 121
8 0
3 years ago
Read 2 more answers
fter production, a computer component is given a quality score of A, B, and C. On the average U of the components were given a q
RoseWind [281]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given question some of the data is missing so, its correct question is defined in the attached file please find it.

Let

A is quality score of A

B is quality score of B

C is quality score of C

\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\

Let F is a value of the content so, the value is:

\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\

Now, we calculate the tooling value:

\to p[\frac{C}{F}]

using the baues therom:

\to p[\frac{C}{F}] =  \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }  

            =  \frac{ 0.17  \times 0.14 }{0.55 \times 0.15 + 0.28 \times 0.12 + 0.17 \times 0.14  } \\\\=  \frac{0.0238}{0.0825 + 0.0336 + 0.0238} \\\\=  \frac{0.0238}{0.1399} \\\\=0.1701

7 0
3 years ago
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