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3 years ago

What is x in x - 4 (x - 3) + 7 = 6 - (x - 4)

Mathematics

1 answer:

WINSTONCH [101]3 years ago

6 0

-1 use mathpapa it will give you the answar

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3. Reasoning Ben says that n = 5 is the solution of the equation 7n = 45. How can you check whether Ben is correct?

likoan [24]

Answer:

You cant, because Ben is WRONG, unless you made a typo while writing the question

Step-by-step explanation:

If you subsitute N for 5, then the equation would be 7 x 5 = 45. However, this is not true. The correct equation is 7 x 5 = 35, so Ben is wrong UNLESS you made a typo while writing the question.

Hope this helped! If it did, please mark it brainliest! It would help a lot! :D

6 0

3 years ago

Problem is in the picture. HELP asap please!! I’m so confused

sashaice [31]

Answer:

equation is

2(x+10x)=1540

width= 70ft

length= 700ft

hope it helps

plz mark brainliest

5 0

3 years ago

Read 2 more answers

2. 90 rock CD's out of 125 CD's.

lidiya [134]

Answer:

72%

Step-by-step explanation:

90/125 is .72

out of 125 CD's there are 72% of rock CD's and 28 of unknown CD's

5 0

3 years ago

Which of the following equations has

Tju [1.3M]

The One With Inifinetely Many Solutions is D. 3x-5=-5+3x are both the same.Meaning that no matter Which Value u Put For X u will get the same Answer On Both Side

7 0

3 years ago

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The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$

7 0

3 years ago

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