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3 years ago

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

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I'm pretty sure it is 3 servings.

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3 years ago

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 years ago

3.01(the 1 is repeating)as a mixed number

Masteriza [31]

The quick and easy answer is 3/100

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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

0.15q=7.8

q=52

d=94-q=42

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