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riadik2000 [5.3K]
3 years ago
7

Suppose y varies directly with x and y= 24 when x=8 what is the value of y when x =10

Mathematics
1 answer:
AlekseyPX3 years ago
5 0

Answer:

B. 30

Step-by-step explanation:

y  ∝  x  or  y  =  k  ⋅  x  or  k  =  \frac{y}{x}  ;   k is constant of variation.  

y  =  24  ;  x  =  8  ∴  k  = \frac{24}{8} =  3  ∴  y  =  3  x.  So When  x  =  10  ;  y  =  3  ⋅  10  =  30

y  =  30

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What is the solution for r in the following system of equations?<br> 2x + 3y = 16<br> 2 + 2y = 10
klio [65]

Answer:

(2,4)

*View attached graph*

Step-by-step explanation:

2x + 3y = 16

2 + 2y = 10

2 + 2y = 10

-2          - 2

2y = 8

/2  /2

y = 4

2x + 3y = 16

2x + 3(4) = 16

2x + 12 = 16

    - 12   - 12

2x = 4

/2   /2

x = 2

(x,y) -> (2,4)

Hope this helps!

8 0
2 years ago
3
katovenus [111]

Question is Incomplete, Complete question is given below.

Mr. Jablonski's class is making fudge for a bake sale. Mr. Jablonski

has a recipe that makes 3/4 pound of fudge. There are 21 students in

the class. Each student uses the recipe to make one batch of fudge.

How many pounds of fudge do the students make?

Answer:

student made \frac{63}{4} \ lbs \ OR \ 15 \frac{3}{4}\ lbs \ OR \ 15.75\ lbs of fudge.

Step-by-step explanation:

Given:

Mr. Jablonski  recipe makes 3/4 pound of fudge.

Number of student in class = 21 students.

Each student uses the recipe to make one batch of fudge.

Hence to calculate Total amount of fudge student made we will multiply the amount of fudge used in recipe with Number of students.

Total amount of fudge student made = amount of fudge used in recipe × Number of students.

Total amount of fudge student made = \frac{3}{4}\times21 = \frac{63}{4} \ lbs =15 \frac{3}{4}\ lbs = 15.75\ lbs

Hence the student made \frac{63}{4} \ lbs \ OR \ 15 \frac{3}{4}\ lbs \ OR \ 15.75\ lbs of fudge.

8 0
3 years ago
Graph the linear equation: - x + 3y = 5
yKpoI14uk [10]
 its easiest if you put it in y = mx + b form, where b is the number you plot on the y axis and m is your slope. 
rearranging... 
x + 3y = -5 3y = -x - 5 y = -1/3 x - 5/3 
so your b value is -5/3, which is where you'll plot a point on the y axis. the slope is -1/3 
since slope is rise over run, [and this case its negative] you'll go down 1 and to the right three. 
edit: then draw a point there, connect the points, and you'll have a line =]
7 0
3 years ago
In ΔXYZ and ΔEFG, angles Y and F are right angles. Which set of congruence criteria would be enough to establish that the two tr
Deffense [45]

XZ ≅ EG and YZ ≅ FG is enough to make triangles to be congruent by HL. Option b is correct.

Two triangles ΔXYZ and ΔEFG, are given with Y and F are right angles.

Condition to be determined that proves triangles to be congruent by HL.

<h3>What is HL of triangle?</h3>

HL implies the hypotenuse and leg pair of the right-angle triangle.

Here, two right-angle triangles ΔXYZ and ΔEFG are congruent by HL only if their hypotenuse and one leg are equal, i.e. XZ ≅ EG and YZ ≅ FG respectively.

Thus, XZ ≅ EG and YZ ≅ FG are enough to make triangles congruent by HL.

Learn more about HL here:

brainly.com/question/3914939

#SPJ1

In ΔXYZ and ΔEFG, angles Y and F are right angles. Which set of congruence criteria would be enough to establish that the two triangles are congruent by HL?

A.

XZ ≅ EG and ∠X ≅ ∠E

B.

XZ ≅ EG and YZ ≅ FG

C.

XZ ≅ FG and ∠X ≅ ∠E

D.

XY ≅ EF and YZ ≅ FG

8 0
2 years ago
Each page of a photo album holds three rows of four square photos. The area of each photo is 25 cm². There is 2 cm space between
Feliz [49]
<span>32 cm wide by 25 cm tall. Since each photo is square and has an area of 25 cm², that means that the length of each edge of the photos is 5 cm. So the length of a page (2 border gaps + 4 photos + 3 spacing gaps) is 2*3 + 4*5 + 3*2 = 32 cm The height of a page (2 border gaps + 3 photos + 2 spacing gaps) is 2*3 + 3*5 + 2*2 = 25 cm So the page size is 32 cm x 25 cm</span>
8 0
3 years ago
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