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3 years ago

12

Jessica is baking a cake. The recipe says that she has to mix 96 grams of sugar into the flour. Jessica knows that 1 cup of this

particular sugar has a mass of 128 grams. She added one over two of a cup of sugar to the flour. Should Jessica add more sugar to make the exact recipe, or did she go over and by what amount?
She went over by two over three of a cup

She needs to add two over three of a cup

She went over by one over four of a cup

She needs to add one over four of a cup

2 answers:

maria [59]3 years ago

4 0

One/four
thats the answer

Elden [556K]3 years ago

3 0

Hello and thanks for posting your question:)

The question is asking if Jessica gave to much or to little to the cake recipe. The answer is She needs to add 1/4 of a cup.

I hope this answer helps. Have a great day:)

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3 years ago

What's 2 meters to 76 centimeters in simplest form

Komok [63]

Answer:

I am not really getting your question. Could you comment on this answer?

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2 years ago

Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert

Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

$s=ut+\frac{1}{2} at^{2}$

Here, s = distance of the balloon above the ground = - 80 feet

u = intital velocity = 16 feet per second

a = acceleration of the object = -32 feet per second

t = required time

So, $s=ut+\frac{1}{2} at^{2}$

$-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})$

$-80=16t-16 t^{2}$

$16 t^{2} -16t -80 =0$

$t^{2} -t -5 =0$

Now, we will use the quadratic D formula for finding the value of t, i.e;

$t = \frac{-b\pm \sqrt{D } }{2a}$

Here, a = 1, b = -1, and c = -5

Also, D = $b^{2} -4ac$ = $(-1)^{2} -(4 \times 1 \times -5)$ = 21

So, $t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}$

$t = \frac{1\pm \sqrt{21 } }{2}$

We will neglect the negative value of t as time can't be negative, so;

$t = \frac{1+ \sqrt{21 } }{2}$ = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

$v=u+at$

Here, v = final velocity

So, $v=16+(-32 \times 2.79)$

v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

8 0

3 years ago

Lucy’s hair grew 5/12 foot over the past year

maks197457 [2]

So about 2 inches or 5 inches

3 0

3 years ago

Read 2 more answers

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ExtremeBDS [4]

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0.6%=0.006 which is $\frac{3}{500}$

I hope this helps.

3 0

3 years ago