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3 years ago

GODD SOMEONE HELP ME RIGHT NOW PLEASE THIS IS DUE TODAY AND I AM DUM I GIVE POINTS AND BRAINLIEST JUST HELP MEEE

Mathematics

1 answer:

Sedbober [7]3 years ago

4 0

Answer:

-5 and -6

Step-by-step explanation:

For trinomial factoring, you need two numbers that add up to get the -11x and two numbers that multiply to get 30.

-5 + (-6) = -11

-5 × -6 = 30

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What is (2w-9) + (-4w-5)

Scrat [10]

Answer:

-2w-14

Step-by-step explanation:

7 0

3 years ago

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How would I start to solve number 15? Would I use distributive property?

natita [175]

$-5-x=2x-(4x+6)$

Using the distributive property is the correct first step. $(4x+6)$ is being multiplied by -1, so distributing gives

$-5-x=2x+(-4x-6)$
$-5-x=2x+((-4x)+(-6))$

Now, using the associative property of addition, you get

$-5-x=(2x+(-4x))+(-6)$
$-5-x=(2x-4x)-6$

Simplify the right hand side:

$-5-x=-2x-6$

Add 5 and $2x$ to both sides.

$(2x+5)-5-x=(2x+5)-2x-6$

On the left side, use the associative property. On the right, commutativity and associativity:

$2x+(5-5)-x=5+(2x-2x)-6$

Simplify.

$2x-x=-1$
$x=-1$

8 0

3 years ago

Find the perimeter of the figure below. Notice that one side length is not given.

sasho [114]

Answer:

58cm

Step-by-step explanation:

15+14+5+8+6=48

missing side = 15-5 =10

48+10=58

3 0

3 years ago

What is<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B81%7D%20" id="TexFormula1" title=" \sqrt{81} " alt=" \sqrt{81} " align

azamat

Answer:

$\: \sqrt{81 } = 9$

Step-by-step explanation:

$\sqrt{81 } = \sqrt{3 \times 3 \times 3 \times 3} \\ = 3 \times 3 \\ =9$

You have to prime factorise 81 and then you will get <u>9</u> as the answer.

8 0

3 years ago

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Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary

Naddik [55]

Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
\\\\\\
AB=-3\pm\sqrt{130}\implies AB=
\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
-3-\sqrt{130}
\end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).

3 0

3 years ago

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