All categories

3 years ago

Amanda used 45 ounces of sugar to make 75 cupcakes. How many ounces of sugar will be used to make 255 cupcakes?

2 answers:

7 0

153oz.

Explanation:

First divide 75 by 45 to get 1⅔. This is the ratio of sugar to cupkakes.

Next take 255 and divide by 1⅔ and you get 153 oz of sugar.

Hope this helps!

kotegsom [21]3 years ago

7 0

Answer:

153 ounces of sugar will be needed.

Step-by-step explanation:

We will use unitary method to calculate the amount of sugar to make 255 cupcakes by Amanda

∵ To make 75 cupcakes amount of sugar needed = 45 ounces

∴ To make 1 cupcake amount of sugar will be needed = 45/75

∴ 255 cupcakes will need amount of sugar = (45/75)×255

= (0.6) × 255

= 153 ounces

Therefore 153 ounces is the answer.

You might be interested in

Given f(x) = 5x-5 find f(-2)

lys-0071 [83]

Answer:

f(-2) = -15

Step-by-step explanation:

f(x) = 5x-5

Let x = -2

f(-2) = 5(-2) -5

= -10 -5

=-15

5 0

3 years ago

In the diagram, lines AB , CD, and EF intersect at G

Ghella [55]

Answer:

D. 70°

Step-by-step explanation:

Given:

m<DGB = 35°

m<CGF = 75°

Required:

m<AGE

SOLUTION:

m<EGD = m<CGF (vertical angles are congruent)

m<EGD = 75°

m<AGE + m<EGD + m<DGB = 180° (angles on a straight line)

m<AGE + 75° + 35° = 180° (substitution)

m<AGE + 110° = 180°

Subtract 110 from each side of the equation

m<AGE = 180° - 110°

m<AGE = 70°

6 0

2 years ago

What is the value of 4(25-5)?

sweet [91]

First you subtract 25 from 5 which equals 20 then you multipliy 4 to 20 and you get 80...so 80 is your solution.

7 0

3 years ago

Help needed ASAP will give BRAINLIEST

grin007 [14]

Answer:

Step-by-step explanation:

Recall that the sum of the (3) interior angles of a triangle <em>must</em> equate to 180.

In other words:

$(55)+(20)+(x)=180$

Where x is our unknown angle. So, we need to solve for x.

On the left, add 55 and 20:

$75+x=180$

Subtract 75 from both sides. The left side cancels:

$x=105$

Therefore, the third angle is 105 degrees.

Our answer is C.

7 0

3 years ago

H(x)=1/8x^3-x^2<br><br> Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$

6 0

3 years ago