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3 years ago

8

The first line in the system of equations is graphed on the coordinate plane. Graph the second line to find the solution to the

system.
2 x minus 3 y = negative 6. y = negative one-third x minus 4
Enjoli’s Graph

2 answers:

Marina CMI [18]3 years ago

4 0

Answer:

(A) (-6,-2)

Step-by-step explanation:

soldier1979 [14.2K]3 years ago

3 0

Answer:

(-6, -2)

Step-by-step explanation:

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What is the value of x enter answer in box

Alona [7]

Answer:

The value of x is 7 cm i think

Step-by-step explanation:

4 0

3 years ago

Solve the following inequality: 3-5x-x^2>=0

Dennis_Churaev [7]

Answer:

-5/2+-1/2√37≤x≤-5/2+1/2√37

Step-by-step explanation:

Step 1: Find the critical points

-x^2-5x+3=0

For this equation: a=-1, b=-5, c=3

−1x^2+−5x+3=0

x=−b±√b2−4ac/2a

x=−(−5)±√(−5)2−4(−1)(3)/2(-1)

x=5±√37 /−2

x=-5/2+1/2√37

Step 2: Check intervals in between critical points

x≤-5/2+1/2 √37 (Doesn't work in original inequality)

-5/2+-1/2√37≤x≤-5/2+1/2√37 (Works in original inequality)

x≥-5/2+1/2 √37 (Doesn't work in original inequality)

5 0

3 years ago

Sal is participating in a charity walk. The distance Sal walks, in kilometers, is proportional to the number of hours he spends

vredina [299]

the attachment is not clear dude...

5 0

3 years ago

HELLLLPPPPPPPP PLEASEEEEEEEEE

tatyana61 [14]

Answer:

10.5

Step-by-step explanation:

$KET\cong KIT...... (given)$

$\therefore KE \cong KI.... (csct)$

$\therefore m(KE) =m(KI)$

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3 years ago

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:

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

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

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3 years ago