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3 years ago

Please help me with this 6 problem please please and I need work to please

Mathematics

1 answer:

klio [65]3 years ago

8 0

What do you need help with? it helps us to know the question!

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The answers To this I don’t understand

Nataly_w [17]

Answer:nvuycghg

Step-by-step explanation:

ngyufuf

6 0

3 years ago

The perimeter of a triangle is 65 meters. The second side is 5 meters less than the first side The third side is 7

tangare [24]

Answer:

The length of each side is 31.5m, 26.5m, 7m

Step-by-step explanation:

Let the length of the first side of the triangle be x meters.

Then, the second side is x-5 meters.

The third side is given as 7 meters.

The perimeter is 65 meters

This gives us the equation:

$65 = x + x - 5 + 7$

$65 - 7 + 5 = x + x$

$63 = 2x$

$x = \frac{63}{2} = 31.5$

The length of each side is 31.5m, 26.5m, 7m

4 0

4 years ago

Can someone turn 4,971 into a scientific notation

Illusion [34]

We need to divide 4,971 by 10 until its a number less then 10.

4,971 / 10 = 497.1

497.1 / 10 = 49.71

49.71 / 10 = 4.971

Now, 4.971 is less then 10, so, we can write this into scientific notation.

4.971 * 10^?

We need to know what power we add to 10 to get 4,971. We can solve for looking at the number of times we divided. As you can see above, we divided 3 times so that will be the power of 10.

4.971 * 10³

Best of Luck!

6 0

3 years ago

Read 2 more answers

Will give brainlliest :>

Goshia [24]

Answer:

48 degrees

Step-by-step explanation:

1) Solve for x

Complementary angles add up to 90 degrees. Therefore, m<A + m<B is equal to 90:

$A+B=90\\(2x+20)+(3x)=90$

Open up the parentheses

$2x+20+3x=90\\5x+20=90$

Subtract 20 from both sides

$5x+20-20=90-20\\5x=70$

Divide both sides by 5

$\frac{5x}{5} =\frac{70}{5} \\x=14$

Therefore, x is equal to 14.

2) Determine angle A

$A=2x+20$

Substitute x with 14

$A=2(14)+20\\A=28+20\\A=48$

Therefore, the measure of angle A in degrees is 48 degrees.

I hope this helps!

3 0

3 years ago

n a particular day, the wind added 2 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 2 miles p

goldfiish [28.3K]

Recall your d = rt, or distance = rate * time

notice, the time he took to cover 63 miles with the wind, is the same amount of time he took against it with 51 miles only, let's say, he took "t" hours long
and he was cycling at a rate of "r"

thus $\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{with the wind}&63&r+2&t\\
\textit{against the wind}&51&r-2&t
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
63=(r+2)t\implies \cfrac{63}{r+2}=\boxed{t}\\\\
51=(r-2)t\\
----------\\
51=(r-2)\left( \boxed{\cfrac{63}{r+2}} \right)
\end{cases}$

solve for "r"

8 0

3 years ago