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3 years ago

If 40 hamburgers feed 30 people how many hamburgers will be needed to feed 90

Mathematics

2 answers:

dezoksy [38]3 years ago

8 0

Answer:120

Step-by-step explanation:

hjlf3 years ago

8 0

Answer:

120 Burgers

Step-by-step explanation:

Your ratio is 40:30 burgers to people. You need to make the 30 into a 90. In other words, multiply by 3 on both sides. 40*3 and 30*3. This gives you a new ratio of 120:90 burgers to people. So take the 120 and you have your answer.

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For what values of x and y must the figure have in order to be a parallelogram

erma4kov [3.2K]

Given:

Given that the parallelogram with the lengths 5x, 3x + 1, 2y - 5 and y.

We need to determine the values of x and y.

Values of x and y:

We know the property that the opposite sides of a parallelogram are congruent.

Thus, we have;

$5x=2y-5$ ------- (1)

$3x+1=y$ ------- (2)

The value of x and y can be determined using the substitution method.

Substituting equation (2) in equation (1), we have;

$5x=2(3x+1)-5$

$5x=6x+2-5$

$5x=6x-3$

$-x=-3$

$x=3$

Thus, the value of x is 3.

Substituting x = 3 in equation (2), we have;

$3(3)+1=y$

$9+1=y$

$10=y$

Thus, the value of y is 10.

5 0

3 years ago

Anyone know the answer? THANK YOU

svetlana [45]

Answer:

The one on the top left

Step-by-step explanation:

Slope of one third and y intercept -4

7 0

3 years ago

Find the value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2

soldier1979 [14.2K]

The value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2 is -2

Solution:

Given that line is passing through point (-5, 2) and (3, r)

Slope of the line is $\frac{-1}{2}$

Need to determine value of r.

Slope of a line passing through point $\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)$ is given by following formula:

$\text { Slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ --- eqn 1

$\text { In our case } x_{1}=-5, y_{1}=2, x_{2}=3, y_{2}=\mathrm{r} \text { and } m=-\frac{1}{2}$

On substituting the given value in (1) we get

$\begin{array}{l}{-\frac{1}{2}=\frac{r-2}{3-(-5)}} \\\\ {\text { Solving the above expression to get value of } r} \\\\ {=>-\frac{1}{2}=\frac{r-2}{3+5}} \\\\ {=>-8=\frac{r-2}{3+5}} \\\\ {=>-8=2(r-2)} \\\\ {=>-8=2 r-4} \\\\ {=>2 r=-8+4} \\\\ {=>2 r=-4} \\\\ {=>r=\frac{-4}{2}=-2}\end{array}$