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Bad White [126]
3 years ago
14

How do I solve this problem 15 = w + 4

Mathematics
2 answers:
vova2212 [387]3 years ago
3 0

Answer:

w = 11

Step-by-step explanation:

<u>Step 1:  Subtract 4 from both sides</u>

15 = w + 4

15 - 4 = w + 4 - 4

11 = w

Answer: w = 11

marissa [1.9K]3 years ago
3 0

Answer:w=11

Step-by-step explanation:

15=w+4

Subtract 4 from both sides

15-4=w+4-4

11=w

w=11

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After the fraction x plus 1 all over 2 minus the fraction x plus 2 all over 3 x have been combined using the least common denomi
RoseWind [281]

If I've read this correctly, it looks like this.

\dfrac{(x + 1)}{2 - \dfrac{(x + 2)}{3x}}

If that is correct, then the first step is to put the top part of the denominator over 3x

\dfrac{(x + 1)}{\dfrac{6x - (x + 2)}{3x}} = \dfrac{(x + 1)}{\dfrac{5x -2}{3x}}

The next part is to flip a three tier fraction. I'm afraid I have to show what happens. My latex is not that strong.

What you get is

\dfrac{3x*(x + 1)}{(5x - 2)}

This is just about your final answer. You could write it as

\dfrac{3x^2 + 3x}{(5x - 2)}

6 0
3 years ago
braden is hiking on a mountain. At 11:00 am he is at an elevation of 500 feet. At 2:00pm he is 900 fett. Find the rate of change
den301095 [7]

Answer:

He changed the elevation by 400 feet

7 0
2 years ago
HELP ME PLEASE IVE BEEN STRUGGLING WITH THIS FOR SO LONG
Ber [7]

Answer:

<h2>m∠ACE = 90°</h2>

Step-by-step explanation:

Figure Interpretation:

m∠CBA + m∠CDE = 180

m∠BCA = (180-m∠CBA)/2

m∠DCE = (180-m∠CDE)/2

=======================

Then

m∠BCA + m∠DCE = (180-m∠CBA)/2 + (180-m∠CDE)/2

                                = [360-(m∠CBA+m∠CDE)]/2

                                = [360 - 180]/2

                                = 90

finally,

m∠ACE= 180 - (m∠BCA + m∠DCE)

             = 180 - 90

             = 90

8 0
2 years ago
Prove that <br><img src="https://tex.z-dn.net/?f=4cot45%20%5E%7B2%7D%20%20-%203tan%20%5E%7B2%7D%2060%20%2B%202sec%20%5E%7B2%7D60
sesenic [268]

Answer:

4 \cot {}^{2} (45)  - 3 \tan {}^{2} (60)  + 2 \sec {}^{2} (60)  = 3 \\

LHS = 4 \cot {}^{2} (45)  - 3 \tan {}^{2} (60)  + 2 \sec {}^{2} (60)  \\

let us first take a look at the values of the trigonometric ratios given in the question so that we get quite clear about what is to be done.

here ,

\cot(45)  = 1 \\  \\  \tan(60)  =  \sqrt{3}  \\  \\  \sec(60)  = 2

now ,

we just have to plug in the values considering certain other things given in the question and we're done!

so let's start ~

4(1) {}^{2}  - 3( \sqrt{3} ) {}^{2}  + 2(2) {}^{2}  \\ \\ \dashrightarrow \: 4(1) - 3(3) + 2(4) \\ \\ \dashrightarrow \: 4 - 9 + 8 \\\\ \dashrightarrow \: 12 - 9 \\ \\\dashrightarrow \: 3 = RHS

hence , proved ~

hope helpful :D

7 0
2 years ago
I need help with this geometry question asap!
guajiro [1.7K]

No pair of lines can be proven to be parallel considering the information given, therefore, the answer is: D. None of the options are correct.

<h3>When are Two Lines Proven to be Parallel to each other?</h3>

Two lines that are cut across by a transversal can be proven to be parallel to each other if:

  • The alternate interior angles along the transversal and on the two lines are congruent [alternate interior angles theorem].
  • The alternate exterior angles along the transversal and on the two lines are congruent [alternate exterior angles theorem].
  • The same-side interior angles along the transversal and on the two lines are supplementary [same-side interior angles theorem].
  • The corresponding angles along the transversal and on the two lines are congruent [corresponding  angles theorem].

Thus, given the following information:

m∠2 = 115°

m∠15 = 115°

With only these two angles given, we can't use any of the theorems to prove that any of the two lines are parallel because angle 2 and angle 15 are located entirely on two different transversals that crosses two lines.

In summary, we can conclude that:

D. None of the options are correct.

Learn more about the Parallel lines on:

brainly.com/question/16742265

#SPJ1

6 0
2 years ago
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