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3 years ago

7

Please need help on this one

2 answers:

jolli1 [7]3 years ago

6 0

30 annanamsmsnskksksksksksk

tresset_1 [31]3 years ago

5 0

If it’s asking for the degrees then the answer is 30, I done that before so..

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How do you solve<br><img src="https://tex.z-dn.net/?f=59%20%3E%202x%20%20%2B%2045" id="TexFormula1" title="59 > 2x + 45" alt

Reika [66]

So you would act like it was an equal sign and solve it like normal. Subtract 45 from both sides. Get 14"="2x. Divide by 2 and get 7 >x. If you were to multiply or divide by a negative than you would flip the sign, but you leave it in this case because you divided by a positive. I hope that helps.

3 0

3 years ago

13. Dave wants to buy a ceramic vase. The sale price is £3.

Irina-Kira [14]

Answer:

0.7x=3

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Hope This Helps!!!

6 0

3 years ago

Point B has coordinates (1,2). The x-coordinate of point A is negative 8. The distance between point A and point B is 15 uni

olchik [2.2K]

Answer:

The point A will be (-8,14) or (-8,-10).

Step-by-step explanation:

Point B has coordinates (1,2) and the x-coordinate of point A is - 8.

Let us assume that the coordinates of point A are (-8,k).

Now, given that the point A is 15 units apart from point B.

Therefore, from the distance formula, we can write that

$\sqrt{(1 - ( -8))^{2} + (2 - k)^{2}} = 15$

Now,squaring both sides, we get

$(1 - ( -8))^{2} + (2 - k)^{2} = 225$

⇒ $(2 - k)^{2} = 225 - 9^{2} = 144$

⇒ 2 - k = ± 12

⇒ k = 14 or -10.

Therefore, the point A will be (-8,14) or (-8,-10). (Answer)

We know that the distance between two points on the coordinate plane ( $x_{1}, y_{1}$ ) and ( $x_{2}, y_{2}$ ) is given by

$\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$ .

7 0

4 years ago

List the numbers that have 30 as a multiple

Yuki888 [10]

30 has Infinite amount of multiples but a few are 60,90,120,150,180,210,240

In simpler terms multiples are what numbers go into

7 0

3 years ago

Approximate √12 to the nearest tenth.

Serjik [45]

Answer:

√12 in nearest tenth is 3.5 is the answer

7 0

3 years ago

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