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2 years ago

10

10. Factor Please help me I will mark as brainliest

1 answer:

suter [353]2 years ago

6 0

Answer:

<u>(3w + 7)² or (3w + 7)(3w + 7)</u>

Step-by-step explanation:

<u>Identity</u>

a² + 2ab + b² = (a + b)² = (a + b)(a + b)

<u>Solving</u>

9w² + 42w + 49
(3w)² + 2(3w)(7) + (7)²
<u>(3w + 7)² or (3w + 7)(3w + 7)</u>

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3 years ago

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3 years ago

Can someone please answer this???? ASAP

timofeeve [1]

Answer:

A. $y=-\dfrac{8}{7}x-\dfrac{18}{7}$

Step-by-step explanation:

Let $y=mx+b$ be the equation of the perpendicular line.

Two perpendicular lines have slopes with product equal to -1. The slope of the given line is $\frac{7}{8}$ Hence,

$\dfrac{7}{8}\cdot m=-1\\ \\m=-\dfrac{8}{7}$

is the slope of needed line.

This line passes through the point (-4,2), so its coordinates satisfy the equation:

$2=-\dfrac{8}{7}\cdot (-4)+b\\ \\b=2-\dfrac{32}{7}=-\dfrac{18}{7}$

Therefore, the equation of the line is

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7 0

3 years ago

Select the expressions that are equivalent to 8(6m).

Bas_tet [7]

$\textsf{Hey there!}$

$\text{Select the expressions that are equicalent to 8(6m)}\downarrow$

$\text{We DON'T have any like terms so we CAN'T pair them off!}$

$\mathsf{8(6m) \rightarrow 8 \times 6 \times \ m}\\\mathsf{8\times 6 = 48}}\\\mathsf{48\times\ m = 48m}\\\\\\\boxed{\boxed{\mathsf{Answer: \bf{48m}}}}\checkmark$

$\textsf{Good luck on your assignment and enjoy your day!}$

~ $\dfrac{\frak{LoveYourselfFirst}}{:)}$

4 0

3 years ago

In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a

nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

$\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ}$ ;

$\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ}$ ;

$\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q$ .

Also,

$\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P$ in right triangle PBM.

Thus $\rm \angle{P\hat{B}M} = \angle C\hat{M}Q$ .

Additionally $\rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}$ .

Therefore $\rm \triangle PBM \sim \triangle MCQ$ .

$\rm \displaystyle BC = 2\;MC$ for M is the midpoint of segment BC.

$\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC$ .

$\rm \triangle PBM \sim \triangle MCQ$ implies that

$\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3$ .

8 0

3 years ago