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2 years ago

Can someone please help me ..I can’t get anymore questions wrong or else I will fail I need help

Mathematics

2 answers:

dedylja [7]2 years ago

5 0

Answer:

maybe number 2

Step-by-step explanation:

because if 2+k=18 then there is no other k there and where did the additional 2 come from ?and 2k=18 then value of k is 9 + there is balance.I may be wrong correct me but i think it is no.2

andrey2020 [161]2 years ago

3 0

Answer:

2k = 18

k = 9

Step-by-step explanation:

First we have to look at each side of the hanger.

There are 2 k's, which can be represented as "2 times the value of k" or 2k.

On the other side, we have the value of 18. The question states that the two sides are balanced. Therefore, the correct equation that represents the image is:

2k = 18.

To find the value of k that makes the equation true, we must isolate the variable k. To do this, we divide 2 from each side of the equation.

2k ÷ 2 = 18 ÷ 2

k = 9.

The value of k that makes the equation true is 9.

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If 15 percent of the sixth graders have blonde hair, and there are 120 sixth graders, how many of them have blonde hair?

Vlada [557]

15/100 = 0.15 x 120 = 18

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3 years ago

Given: AB || DE , AD bisects BE. Prove: ABC = DEC using the ASA postulate.

Ket [755]

Answer:

As per ASA postulate, the two triangles are congruent.

Step-by-step explanation:

We are given two triangles:

$\triangle ABC$ and $\triangle DEC$ .

AD bisects BE.

AB || DE.

Let us have a look at two properties.

1. When two lines are parallel and a line intersects both of them, then alternate angles are equal.

i.e. AB || ED and $\angle B$ and $\angle E$ are alternate angles $\Rightarrow$ $\angle B = \angle E$ .

2. When two lines are cutting each other, angles formed at the crossing of two, are known as Vertically opposite angles and they are are equal.

$\Rightarrow \angle ACB = \angle DCE$

Also, it is given that AD bisects BE.

i.e. EC = CB

1. $\angle B = \angle E$

2. EC = CB

3. $\angle ACB = \angle DCE$

So, we can in see that in $\triangle ABC$ and $\triangle DEC$ , two angles are equal and side between them is also equal to each other.

Hence, proved that $\triangle ABC$ $\cong$ $\triangle DEC$ .

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2 years ago

What does no gains or no losses on first down meaning

MissTica

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2 years ago

CAN SOMEONE PLEASE ANSWER THIS SOON! THANK YOU!

lyudmila [28]

We want to find one-half of the reciprocal of 7/sqrt(98). Let's write down an expression for this:

$\dfrac{1}{2} \times \dfrac{\sqrt{98}}{7}$

We can rewrite 98 into $2 \times 49$

$\dfrac{1}{2} \times \dfrac{\sqrt{2 \times 49}}{7}$

$=\dfrac{1}{2} \times \dfrac{\sqrt{2} \times \sqrt{49}}{7}$

The square root of 49 is 7

$=\dfrac{1}{2} \times \dfrac{\sqrt{2} \times 7}{7}$

$=\dfrac{1}{2} \times\sqrt{2}$

$=\dfrac{\sqrt{2}}{2}$

This should be your answer. Let me know if you need any clarifications, thanks!

5 0

2 years ago

Read 2 more answers

Verify cosa/1+sina +1+sina/cosa= 2sec a

Dmitry [639]

Cosa/1+sina + 1+sina/cosa = (cos^2a+1+2sina+sin^2a)/(cosa)((1+sina)
=( 1+1+2sina)/(cosa) (1+sina)
=2(1+sina)/ (cosa) (1+sina)
=2/cosa
=2sec a (verified)

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3 years ago