All categories

3 years ago

I’m don’t know can somebody help me out?

Mathematics

2 answers:

BabaBlast [244]3 years ago

8 0

Answer:

Step-by-step explanation:

30/2=15

Leno4ka [110]3 years ago

4 0

Answer: B

Step-by-step explanation:

2 times 15 = 30 so 1 time 15 = 15

I may be wrong. Hope this helped!!!

You might be interested in

18. Emma rents a car from a company that rents cars by the hour. She has to pay an initial fee of $75, and then they charge her

love history [14]

Step-by-step explanation:

Let h be the number of hours.

Total Cost = Fixed Costs + Variable Cost

= Initial Fee + Hourly Charge

= 75 + 9h

Given,

$75 + 9h = 250 \\ 9h = 250 - 75 \\ 9h = 175 \\ h = 175 \div 9 \\ = 19 \frac{4}{9} hr$

Since the car cannot be rent part of an hour, the highest possible whole number is 19 hours.

5 0

3 years ago

The distance between -18 and 15 is equal to_______

ArbitrLikvidat [17]

Answer:

sorry I forgot try googling it

3 0

3 years ago

---8 (-5 + 13) + 12 : 6 x 2

madreJ [45]

Step-by-step explanation:

-8 ( - 5 + 13) + 2 : 1 x 2

-8 ( 8) + 2 : 2

-64 + 1

-63

6 0

3 years ago

If cos(θ)=2853 with θin Q IV, what is sin(θ)?

marysya [2.9K]

Answer: $\sin \theta=\frac{-45}{53}$

Step-by-step explanation:

Since we have given that

$\cos\theta=\frac{28}{53}$

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

$\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}$

It must be negative due to its presence in Fourth quadrant.

Hence, $\sin \theta=\frac{-45}{53}$

7 0

3 years ago

Need the a answer for this

FinnZ [79.3K]

Answer:

Given expression:

$\dfrac{14a^4b^6c^{-10}}{8a^{-2}b^3c^{-5}}$

Separate the variables:

$\implies \dfrac{14}{8} \cdot \dfrac{a^4}{a^{-2}} \cdot \dfrac{b^6}{b^3} \cdot \dfrac{c^{-10}}{c^{-5}}$

Reduce the first fraction:

$\implies \dfrac{7}{4} \cdot \dfrac{a^4}{a^{-2}} \cdot \dfrac{b^6}{b^3} \cdot \dfrac{c^{-10}}{c^{-5}}$

$\textsf{Apply Division Property of Exponents rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies \dfrac{7}{4} \cdot a^{4-(-2)} \cdot b^{6-3} \cdot c^{-10-(-5)}$

$\implies \dfrac{7}{4} \cdot a^{6} \cdot b^{3} \cdot c^{-5}$

$\textsf{Apply Negative Property of Exponents rule} \quad a^{-n}=\dfrac{1}{a^n}$

$\implies \dfrac{7}{4} \cdot a^{6} \cdot b^{3} \cdot \dfrac{1}{c^5}$

Therefore:

$\implies \dfrac{7a^6b^3}{4c^5}$

4 0

1 year ago