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3 years ago

5

What is the value of x? A) 10 B) 12 C) 15 D) 17

1 answer:

prohojiy [21]3 years ago

7 0

Answer:

c) 15

Step-by-step explanation:

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When 50% of a number is added to the number, the result is 210. what is the number?

stellarik [79]

I hope this helps you

4 0

3 years ago

Read 2 more answers

Can someone help me with this question for my homework 2x+29+11=x+28

givi [52]

Sure! I'll try!

2x + 29 +11 = x + 28

First, you would put like terms ( the x ) on the same side.

2x + 29 + 11 = x + 28

-2x -2x

Giving you: 29 + 11 = -3x + 28

Then, you would add 29 and 11 together to get 40.

Next, you'd put the 28 on the other side.

40 = -3x + 28

-28 - 28

Giving you: 12 = -3x

5 0

3 years ago

Read 2 more answers

If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

4 0

4 years ago

*****WILL GIVE BRAINLEST IF CORRECT******

PSYCHO15rus [73]

Conner's work is correct. To combine and make it simple, you Multiply:

(3^5+9)+(6^8+10) which will equal 3^14 6^18.

But Jane's work, instead of adding, Jane multiplies. So, Conner is correct.

Hope that helped!

5 0

3 years ago

Find the value of the variable.<br><br> m<BDK = 3x+ 4, m<JDR = 5x - 10

slega [8]

∠BDK = ∠JDR
3x +4 = 5x -10
14 = 2x . . . . . . . . add 10-3x
7 = x . . . . . . . . . . divide by 2

5 0

3 years ago