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3 years ago

Help plz due now will mark brainliest if correct

Mathematics

1 answer:

Maurinko [17]3 years ago

3 0

Answer:

2^3

Step-by-step explanation:

you subtract the exponents if they have the same coefficient when dividing them.

9-6=3

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Use synthetic division to find P (-10) for P(x)=2x^3+14x^2-58x

Savatey [412]

The polynomial remainder theorem states that the remainder upon dividing a polynomial $p(x)$ by $x-c$ is the same as the value of $p(c)$ , so to find $p(-10)$ you need to find the remainder upon dividing

$\dfrac{2x^3+14x^2-58x}{x+10}$

You have

..... | 2 ... 14 ... -58
-10 | ... -20 ... 60
--------------------------
..... | 2 ... -6 .... 2

So the quotient and remainder upon dividing is

$\dfrac{2x^3+14x^2-58x}{x+10}=2x-6+\dfrac2{x+10}$

with a remainder of 2, which means $p(-10)=2$ .

5 0

3 years ago

Use the law of sines to find the value of w.

Lunna [17]

Option B: 4.0 cm is the value of w

Explanation:

It is given that $\angle W= 39$ , $\angle U=31$ , $VW=3.3$ and $UV=w$

Using the law of sines, we have,

$\frac{sin U}{3.3} =\frac{sin W}{w}$

Substituting the values, we have,

$\frac{\sin 31}{3.3}=\frac{\sin 39}{w}$

Cross multiplying, we get,

$(sin31) w=3.3( \sin 39)$

Simplifying, we have,

$0.515w=2.077$

Dividing both sides by 0.515, we have,

$w=4.0$

Thus, the value of w is 4.0 cm

Hence, Option B is the correct answer.

4 0

2 years ago

Read 2 more answers

David & Brad are competing in a 20-mile bicycle race. David rides 6.2 miles in 1/4 hour. Brad rides 11.9 miles in 1/2 hour.

Elodia [21]

David will finish first, because if you divide Brad's miles by 2 to his miles in a 1/4 of an hour are 5.95 miles. This means David went 6.2 miles in the same time that Brad went 5.95. Therefore, David will finish first.

6 0

3 years ago

In a right triangle abc, cd is an altitude, such that ad=bc. Find ac, if ab=3 cm, and cd= square root 2 cm.

Mashcka [7]

AC² + BC² = AB² AD² + CD² = AC²

BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)

BC² = AC² - CD²

AB² - AC² = AC² - CD² (both sides were = to BC²)

AB² + CD² = 2AC²

(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)

9 + 2 = 2AC²

11 = 2AC²