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Lady bird [3.3K]
3 years ago
15

Solve each equation & show 3) 36 = 1+ 7a

Mathematics
2 answers:
Lisa [10]3 years ago
4 0

Answer:

a=5

Step-by-step explanation:

36= 1+ 7a

subtract 1 on both sides

35= 7a

divide by 7 on all sides

5=a

a = 5

Sladkaya [172]3 years ago
3 0

Answer:

a=5

The drawing will help

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The mean of a population being sampled is 64, and the standard deviation is 6. If the sample size is 50, the standard error of t
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3 years ago
Help please math ..??
astraxan [27]

Answer:

(3, - 1)

Step-by-step explanation:

5x - 4y = -11

2x + 3y = -9

Multiply the top equation by 2 and the bottom equation by - 5 to cancel out the x's.

10x - 8y = - 22

- 10x - 15y = 45

Cancel out x's.

-8y = - 22

- 15y = 45

Add like terms.

- 23y = 23

Can't have a negative variable to flip them.

-23 = 23y

Divide.

y = - 1

Input y into one of the equations and solve for x.

2x + 3(-1) = -9

Simplify.

2x -3 = -9

Cancel out -3 by adding 3.

2x = -6

Divide.

x - -3

8 0
3 years ago
A hot air balloon descends from an altitude of 2,000 feet at a constant rate of 90 feet per minute. The graph shows the altitude
liberstina [14]

Answer:

y=90x+2,000

Step-by-step explanation:

90 is the rate of change because the hot air balloon descends 90 feet per minute.

And 2,000 is the starting point because the hot air balloon started the descent at 2,000 feet.

6 0
3 years ago
Read 2 more answers
On the box and whisker plot below, what data calculations do the points B and D represent?
daser333 [38]

Answer:

Lower and upper quartiles

Step-by-step explanation:

The box starts at lower quartile, ends at upper quartile

5 0
3 years ago
The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-
Elena-2011 [213]

y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)

3 0
1 year ago
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