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scoundrel [369]
2 years ago
15

The ratio of students to teachers at a school is 19:1 how many students are there in each total of people?

Mathematics
2 answers:
omeli [17]2 years ago
8 0

Answer:

760  or  D

BRAINLIEST PLZZZZZZZZZZ

Step-by-step explanation:

Valentin [98]2 years ago
3 0

Answer:

D

Step-by-step explanation:

19*40= 760

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I posted a question similar to this but I entered it wrong too many times :(
slava [35]

Answer:

\sqrt[4]{\frac{5}{7} }

Step-by-step explanation:

^4√5/^4√7

~Apply radical rules

^4√5/7

Best of Luck!

5 0
3 years ago
Read 2 more answers
228 + 6 =<br> R<br> What is the quotient
cricket20 [7]

228 + 6 = 224 + 6 + 4 = 230 + 4

= 234

Answer: 234

ok done. Thank to me :>

4 0
2 years ago
3.) You just deposited $3,000 into a savings account with 3% annual interest. How much will you have in total after 2 years?​
Korvikt [17]

Answer:2,820

Step-by-step explanation:

multiply 3,000 by 0.03 which will get you 90 which is 3% of that for one year and for two years you just multiply 90 by 2 and get 180. subtract 180 from 3,000

6 0
3 years ago
What is the value of 2x4 + 5x when x=6?
wlad13 [49]
This would be 2(6)^4 + 5*6

=  2592 + 30

= 2622 Answer
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=&#10;\begin{cases}&#10;x=7cos(330^o)\\&#10;\qquad 7\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{7\sqrt{3}}{2}\\&#10;y=7sin(330^o)\\&#10;\qquad 7\cdot -\frac{1}{2}\\&#10;\qquad -\frac{7}{2}&#10;\end{cases}\qquad \qquad v=&#10;\begin{cases}&#10;x=8cos(30^o)\\&#10;\qquad 8\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{8\sqrt{3}}{2}\\&#10;y=8sin(30^o)\\&#10;\qquad 8\cdot \frac{1}{2}\\&#10;\qquad 4&#10;\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)&#10;\\\\\\&#10;\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\&#10;-------------------------------

\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}}&#10;\\\\\\&#10;\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
3 years ago
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