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bazaltina [42]
3 years ago
13

<A and <J are complementary angles. Find the value of x m<A = 5x + 30 m<J = 4x +15

Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0

Sum of complementary angles always equal to 90°.

Given, <A and <J are complementary angles. So, we can set up an equation as following:

m<A+ M<J= 90

Next step is to plug in m<A = 5x + 30 and m<J = 4x +15 in the above equation. So we will get,

5x+30+4x+15=90

(5x+4x)+(30+15)=90 Group the like terms.

9x+45=90 Combine the like terms.

9x+45-45=90-45 Subtract 45 from each sides.

9x=45 By simplifying.

\frac{9x}{9}=\frac{45}{9} Divide each sides by 9.

x=5.

So, x=5.

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2. Determine the point of intersection for the equations below without graphing. Show all work.
Murljashka [212]
I’m pretty sure you would have to make them equal
1. 3x -10=7x+14
2. Add 10 to both sides
3. 3x=7x+24
4. Subtract the variable ( 7x ) from both sides
5. -4x=24
6. Divide -4/24
I hope I did it right let me know if any errors
8 0
3 years ago
A pen cost y dollars,how much will 25 such pens cost
kiruha [24]

Answer:

25y

Step-by-step explanation:

y×25=25y

3 0
3 years ago
george wanted to measure the angle of a waterslide at the lake .he used a sheet of folded paper that formed 20 degree angle. he
evablogger [386]

Answer:

It's not clear on what it means but the information given it would just be 20*3=60

Step-by-step explanation:

4 0
3 years ago
S
Scilla [17]

0.03

Step-by-step explanation:

S ∝ t, introduce constant, if value of S depends on t only. Let that constant be 'k'.

=> S = kt

When, S = 0.6 & t = 4:

Answer

=> 0.6 = k(4)

=> 0.6/4 = k

=> 0.15 = k

Therefore, when t = 0.2

=> S = kt

= (0.15)(0.2)

= 0.03

3 0
2 years ago
Karen notices that segment BC and segment EF are congruent in the image below:
LenaWriter [7]

Answer:

Givens:

  • A(-2;2) \ B(-2;4) \ C(0;3) \ D(2;1) \ E(2;3) \ F(0;2)
  • BC \cong EF
  • \angle C \cong \angle F

Using this information we could recur to SAS postulate to demonstrate the congruence of this pair of triangles. We already have one side and one angle congruent, we just need to demonstrate that AC \cong DF, that will prove the complete congruence.

So, based on the given coordinates, we're able to calculate the length of each side and find if they are the same. We use this formula:

d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\d_{AC} =\sqrt{(0-(-2))^{2}+(3-2)^{2}}=\sqrt{4+1}=\sqrt{5}\\\\\d_{DF} =\sqrt{(0-2)^{2}+(2-1)^{2}}=\sqrt{4+1}=\sqrt{5}

Therefore, AC \cong DF, because they have the same length.

Now, based on these congruences:

  • BC \cong EF
  • \angle C \cong \angle F
  • AC \cong DF

We demonstrate by the Side-Angle-Side (SAS) postulate that ΔABC ≅ ΔDEF.

6 0
3 years ago
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