All categories

3 years ago

If we know that 5 cups of flour is used to make 20 biscuits, figure out how many biscuits are made with 15 cups of flour

Mathematics

1 answer:

aleksley [76]3 years ago

6 0

Answer:

100

Step-by-step explanation:

5×4=20

5×15=100

therefore 15 cups of flour can make 100 biscuits

You might be interested in

A un tanque que contiene 150 litros de gas propano se le inyecta del mismo gas a razón de 3 litros por segundo. Determina la fun

ElenaW [278]

If only it was in english i could of read it and gave you the answer

7 0

4 years ago

Given the diagram below, find X. The diagram IS 40° x - 15° X = degrees.

Crank

Answer:

X=65°

Step-by-step explanation:

Both angles combined is a right angle as indicated by the small square drawn on it, a right angle is always 90° therefore

(x-15°)+40°=90°

combine like terms

x+25°=90°

subtract 25° from each side

x=65°

6 0

3 years ago

Answer in yd! need help asap

sweet [91]

Answer:

8 yd

Step-by-step explanation:

The opposite side is 14 and the lower portion plus the missing value would equal 14:

14 = 6 + ?

So subtract 6 from 14 to get the missing length, 8.

8 0

3 years ago

Simplify the following (1 1/2) ^-2

Serjik [45]

Answer:

1/2.25 or 4/9

Step-by-step explanation:

6 0

3 years ago

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$

4 0

2 years ago