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2 years ago

Factor the polymonial completely using the X method x2 + 13x - 48 quickest response gets braniliest

Mathematics

1 answer:

rusak2 [61]2 years ago

5 0

Answer:

The factored form comes from breaking each term up x2 = x times x

-48 has numerous possibilities for two factors but the two factors that multiply and get -48 while adding to get +13 are +16 and -3 so the answer in factored form is (x + 16) (x -3)

Step-by-step explanation:

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3 years ago

Divide.<br><br> 7 9/6 divided by 2 3/4

rosijanka [135]

Answer:

2 20/69

Step-by-step explanation:

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2 years ago

Read 2 more answers

Write the monomial in standard form. name it's coefficient and identify its degree.

Hunter-Best [27]

Answer:

$Standard\ Form = {3n^2} m^{-2}$

Step-by-step explanation:

Given

$\frac{2}{3m^2n} * 4.5n^3$

Required

Write in Standard Form

To start with; the two monomials have to be multiplied together;

$\frac{2}{3m^2n} * 4.5n^3$

$Standard\ Form = \frac{2 * 4.5n^3}{3m^2n}$

Split the numerator and the denominator

$Standard\ Form = \frac{2 * 4.5 * n^3}{3 * m^2 * n}$

Multiply Like terms

$Standard\ Form = \frac{9 * n^3}{3 * m^2 * n}$

Divide 9 by 3 to give 3

$Standard\ Form = \frac{3 * n^3}{m^2 * n}$

Divide n³ by n to n²

$Standard\ Form = \frac{3 * n^2}{m^2 }$

Split fraction

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$

From laws of indices;

$\frac{1}{a^n} = a^{-n}$

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$ becomes

$Standard\ Form = {3 * n^2} * m^{-2}$

Multiply all together

$Standard\ Form = {3n^2} m^{-2}$

5 0

3 years ago

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

What are RQS and TQS?

yaroslaw [1]

Answer:

m∠RQS = 72°

m∠TQS = 83°

Step-by-step explanation:

m∠RQS +m ∠TQS = m∠RQT

The two angles combine to make a larger angle

m∠RQS = (4x - 20)

m∠TQS = (3x + 14)

(4x - 20) + (3x + 14) = 155

Group the Xs and the constants

4x + 3x - 20 + 14 = 155

Combine like terms

7x - 6 = 155

Add 6 to both sides

7x = 161

Divide by 7 on both sides

x = 23

Check:

4(23) - 20 + 3(23) + 14 = 155

92 - 20 + 69 + 14 = 155

155 = 155

But we need to find m∠RQS and m∠TQS. So plug in x = 23 to the values.

m∠RQS = 4(23) - 20 = 72°

m∠TQS = 3(23) + 14 = 83°

Checking:

72 + 83 = 155

8 0

3 years ago