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DiKsa [7]
3 years ago
9

Help pls 11 points! Plss

Mathematics
1 answer:
mafiozo [28]3 years ago
5 0

I Think it is D

hope this help


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A large bag of cashews weighs 8 1/3 pounds. One serving is1/3 pound.how many servings are in the bag
denis23 [38]

I'm pretty sure it is 3 servings.


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A delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is
BARSIC [14]
Large boxes are 40 pounds.
Small boxes are 30 pounds.
Let me know if you need the work!
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3.01(the 1 is repeating)as a mixed number
Masteriza [31]
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7 0
3 years ago
When Julio cracked open his piggy bank, he
jok3333 [9.3K]

Answer:

52 quarters, 42 dimes

Step-by-step explanation:

Let's call the number of quarters q, and the number of dimes d.

q+d=94

0.25q+0.1d=17.20

Subtract q from both sides of the first equation to isolate d:

d=94-q

0.25q+0.1(94-q)=17.20

0.25q+9.4-0.1q=17.20

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q=52

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Hope this helps!

5 0
3 years ago
Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30
Dafna11 [192]
Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}

\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------

\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
2 years ago
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