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just olya [345]
3 years ago
6

Which answer is it? :/

Mathematics
1 answer:
telo118 [61]3 years ago
8 0

Answer:

D

Step-by-step explanation:

f(x)=-3x-1

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Your test scores are 75, 93, 90, 82 and 85. What is the lowest score you can obtain on the next test to achieve an average of at
Korvikt [17]
You’ll need a total of 86x6= or 516 total points to meet your average of at least 86. Now subtract out your current scores: 516- 75-93-90-82-85 = 91. You’ll need 91 to meet the 516 total requirement.
5 0
3 years ago
Jenson has 256 beads. He gets 67 more beads from a friend. He then uses 157 of the beads to make necklaces for his teachers. Fin
Murljashka [212]
So Jenson has 256 beads + 67 beads =  323 beads - 157 beads for the <span>necklaces = </span> 166 beads divided by 8 beads per bracelet = 20 bracelets with the remainder of 6 beads.  
8 0
3 years ago
Read 2 more answers
Solve this equation, show all your work:<br> 6(x+1)=-2(3x+9)
Alex17521 [72]

Answer:

x=-2

Step-by-step explanation:

6(x+1)=-2(3x+9)

Distribute the 6.

6x+6=-2(3x+9)

Distribute the -2.

6x+6=-6x-18

Move the variable to the left-hand side and change the sign.

6x+6+6x=-18

Collect like terms

12x=-18-6

Calculate the difference.

12x=-24

Divide both sides by 12.

x=-2

Hope this helps :)

7 0
2 years ago
On Monday at work, David produces $w$ widgets per hour, and works for $t$ hours. Exhausted by this work, on Tuesday, he decides
nasty-shy [4]

Answer:

8 more widgets

Step-by-step explanation:

<u>Monday:</u>

t hours work AT w widgets per hour

So,

total number of widgets = wt

<u>Tuesday:</u>

2 fewer hours than monday, so

hours worked = t - 2

Per hour widget production is 4 more than Monday so:

per hour widget = w + 4

Total number of widgets = (t-2)(w+4)=wt+4t-2w-8

Monday he produced more, so amount more is:

wt-[wt+4t-2w-8]\\wt-wt-4t+2w+8\\-4t+2w+8

Given, w = 2t, we have:

-4t+2w+8\\-4t+2(2t)+8\\-4t+4t+8\\=8

Thus,

8 more widgets were produced

5 0
3 years ago
Prove that angle abc is a right angle with vertices a(-4,3), B(0,-1) and C(2,1)​
spayn [35]

Check the picture below.

clearly the angle at vertex C isn't the right angle, so let's check if AB ⟂ BC.

\bf A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{0}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-3}{0-(-4)}\implies \cfrac{-4}{0+4}\implies \cfrac{-4}{4}\implies -1 \\\\[-0.35em] ~\dotfill

\bf B(\stackrel{x_1}{0}~,~\stackrel{y_1}{-1})\qquad C(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-1)}{2-0}\implies \cfrac{1+1}{2}\implies \cfrac{2}{2}\implies 1 \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{so is a right triangle}}{AB\perp BC}~\hfill

7 0
3 years ago
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