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3 years ago

How do you do 50/4x when x is 10?

2 answers:

7 0

When a number is in front of a variable, it means we have to multiply. We can substitute the x for a 10, since that is what the problem gave us:

50/(4)(10)
=50/40
=5/4

dlinn [17]3 years ago

4 0

4x=40 so 50/40 you got it from there

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The first figure is dilated to form the second figure.

coldgirl [10]

This is clearly a reduction meaning the copy got smaller. This indicates that the scale factor IS less than 1 and the only answer choice less than 1 is the first statement.

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Answer : The scale factor is 0.25.

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7 0

3 years ago

Read 2 more answers

A grocer sells milk chocolate at $2.50 per pound, dark chocolate at $4.30 per pound, and dark chocolate with almonds at $5.50 pe

rosijanka [135]

Answer:

10 pounds of milk chocolate, 15 pounds of dark chocolate, and 10 + 15 = 25 pounds of almond chocolate

Step-by-step explanation:

Let x be the number of pounds of milk chocolate that he needs to use in the mixture. And y is the number of pounds of of dark chocolate. Since the mixture must use almonds chocolate as much as the other 2 combined in term of weight, and the total weight is 50 pounds. That means

milk_choco_weight + dark_choco_weight + almond_choco_weight = 50

x + y + (x + y) = 50

2(x+y) = 50

x+y = 25 or x = 25 - y

Since we know the price of the mixture, we can use the following equation

2.5milk_choco_weight + 4.3dark_choco_weight + 5.5almond_choco_weight = 4.54total_weight

2.5x + 4.3y + 5.5(x+y) = 50*4.54

2.5x + 4.3y + 5.5*25 = 227

2.5x + 4.3y = 227 - 137.5 = 89.5

We can substitute x = 25 - y

2.5(25 - y) + 4.3y = 89.5

62.5 - 2.5y + 4.3y = 89.5

1.8y = 27

y = 15

so x = 25 - y = 25 - 15 = 10

So we need 10 pounds of milk chocolate, 15 pounds of dark chocolate, and 10 + 15 = 25 pounds of almond chocolate

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2 years ago

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2 years ago

Mr. Moore has 21 feet of wallpaper he cuts it into sections that are each 3 feet long how many

9966 [12]

Total = 21 feet

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3 years ago

Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that

RUDIKE [14]

The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)= $f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........$

Where the terms in f prime $f^{1}$ (a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

$f^{1}$ (a)= $f^{1}$ (-4)=-8/( $-4)^{2}$

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+ $f^{1}(-4)(x+4)/1!+ f^{11}(-4)(x+4)^{2}/2!.............$

$=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+ 24/(-4)^{3} (-4)(x+4)^{2}/2!.............$

=-2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Learn more about taylor series at brainly.com/question/23334489

#SPJ4

3 0

1 year ago