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2 years ago

15

Jake bought some blueberries for $6 per pound and some strawberries for $4 per pound. He spent a total of $36. Which graph repre

sents all the possible combinations of blueberries (b) and strawberries (s) that Jake could have bought?

1 answer:

Zarrin [17]2 years ago

7 0

Answer:

Step-by-step explanation:

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pochemuha

Answer:

a =74,321

b. 41,042,523,288

c. 414,832,917

Step-by-step explanation:

because two odd numbers becomes a even number

5 + 5 = 10 7 + 7 = 14 etc

so in the problem you sent 13 and 15 being odd numver will result in a even number when added

8 0

3 years ago

Read 2 more answers

Find the perimeter area and area of the polygon.

Alika [10]

The area is 76 ft squared, and the perimeter is 43.1 ft. CORRECT ME IF IM WRONG BECAUSE I WAS SO BAD AT AREA/PERIMETER

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2 years ago

How do you solve 235+123 using a number line?

Scrat [10]

You would either start at 0 and go over 235 then you would go over 123 more

6 0

2 years ago

All sides of ABC are tangent to circle P.<br> What is the perimeter of the triangle below?

arlik [135]

I got an answer of 58.

Subtract 10 from the side length of 24 to get 14.

And use the side length of 5 to find another side length that also equals 5.

In total all your side lengths would be 10,5,5,24,14. And all of them together to arrive at an answer of 58.

6 0

2 years ago

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

4 0

3 years ago