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3 years ago

The length of the midsegment of a triangle is always half the length of the side it is parallel to true or false

1 answer:

3 0

True. The midsegment of a triangle is halfway up the triangle, and since the top half and the full triangle are similar triangles, its dimensions are proportional.

This means half the height will give half the base (midsegment). So the midsegment is always half the length of the side it's parallel to.

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Consider the function f(x)=-2/3x +5 what is f(-6) enter your answer in the box f(-6)=

GaryK [48]

Answer:

f(- 6) = 9

Step-by-step explanation:

to evaluate f(- 6), substitute x = - 6 into f(x)

f(- 6) = - $\frac{2}{3}$ × - 6 + 5

= - $\frac{2(-6)}{3}$ + 5 = 4 + 5 = 9

4 0

3 years ago

Pls answer take ur time

bekas [8.4K]

Answer:

y = 3/4x

Step-by-step explanation:

We need to find the slope of the line

We have point (0,0) and point (4,3)

m = (y2-y1)/(x2-x1)

= (3-0)/(4-0)

=3/4

The y intercept is 0 since it crosses the y axis at 0

The slope intercept form of the line is

y = mx+b

y = 3/4x +0

y = 3/4x

5 0

3 years ago

Nathan's lawn mower uses 23 of a tank of gas to cut three one-acre lawns. How many one-acre lawns can he cut with a full tank of

Angelina_Jolie [31]

60 Gallons To FInish The Whole Entire Yard I think double check to make sure

6 0

3 years ago

A bakery sells bagels at two different rates. When a customer buys fewer than 12 bagels, each one costs $1.25. When a customer b

Sav [38]

Answer:

1.25b when b < 12 , 1b & then 0.75 b when b > 12

Step-by-step explanation:

Price for b bagels = Price x Quantity

* Less than 12 bagels bought, cost $1.25 per bagel

So, price for b bagels, when (b < 12) = 1.25b

* 12 or more than 12 bagels bought, cost $1 upto first 12 bagels & 0.75b on each additional bagel

So, price for b bagels, when (b = 12) = b

So, price for b bagels, when (b > 12) = b + 0.75b' , where b' denotes extra b beyond 12

8 0

3 years ago

Please help me with this question!

gavmur [86]

First off, let's convert the percentages to decimal format, so our 77% turns to 77/100 or 0.77, and our 55% turns to 55/100 or 0.55 and so on

now, the sum of both salines, must add up to the 77% mixture, let's say is "y"
so, 11 + 4 = y, and whatever the concentration level is, must also sum up to the mixture's concentration of 77%

anyway thus

$\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentrated\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{first sol'n}&11&x&11x\\
\textit{second sol'n}&4&0.55&2.20\\
------&-----&-------&-------\\
mixture&y&0.77&0.77y
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
11+4=y\implies 15=\boxed{y}\\
11x+2.2=0.77y\\
----------\\
11x+2.2=0.77\cdot \boxed{15}
\end{cases}$

solve for "x"

6 0

3 years ago