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4 years ago

Help me with math 20 points

Mathematics

1 answer:

irakobra [83]4 years ago

7 0

I would suggest that you look up a conversion chart.

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3 years ago

If N = 9 ft, Q = 2 ft, R = 4 ft, S = 9 ft, and T = 3 ft, what is the perimeter of the object?

lana66690 [7]

The perimeter of the object NQRST will be 27 ft.

<h3>What is a regular figure and its perimeter?</h3>

A regular figure with n-sides has n equal sides in it, and they are the only parts of it(that means, nothing more than those equal lengthed n sides).

Suppose that length of each side of that figure be of u units, then we have the perimeter as: A + B + C + D units.

Its the sum of length of the sides used to made the given figure.

Given; If N = 9 ft, Q = 2 ft, R = 4 ft, S = 9 ft, and T = 3 ft, then we need to find the perimeter of the object.

So the perimeter of the object NQRST.

N + Q +R + S +T

= 9 + 2 +4 + 9 + 3

= 27 ft

Therefore, the perimeter of the object NQRST will be 27 ft.

Learn more about perimeter here:

brainly.com/question/10466285

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7 0

2 years ago

Eva and Dulcina have 82 pins together. Dulcina has 34 more pins than Eva. How many pins are in Dulcinas collection? How many pin

ludmilkaskok [199]

Answer:

Eva has 24 pins

Step-by-step explanation:

D = 34 + E

E + D = 82

34 + E + E = 82

2E = 48

E = 24

D = 24 + 34

D = 58

7 0

3 years ago

Jason practiced making soccer goals and kept her track of the results. He missed 3 goals but made 9 in one practice session. Wha

PIT_PIT [208]

He can predict to make 8 out of 24

5 0

3 years ago

A figure is broken into a rectangle and a triangle. The triangle has a base of 2 and two-thirds feet and height of 3 feet. The r

ohaa [14]

Answer:

$b=2\frac23\:\sf ft$

Area of triangle = 4 ft²

Area of rectangle = $6\frac23\: \sf ft^2$

Area of irregular figure = $10\frac23\: \sf ft^2$

Step-by-step explanation:

$\begin{aligned}\sf Base\:of\:triangle\:(b) & = 5-2\frac13\\\\& = \dfrac{15}{3}-\dfrac73\\\\ & = \dfrac83\\\\ & = 2\frac23\:\sf ft \end{aligned}$

$\begin{aligned}\sf Area\:of\:a\:triangle & =\dfrac12 \sf \times base \times height\\\\& = \dfrac12 \times b \times 3\\\\ & = \dfrac12 \times \dfrac83 \times \dfrac31\\\\ & = \dfrac{24}{6}\\\\ & = 4\: \sf ft^2\end{aligned}$

$\begin{aligned}\sf Area\:of\:rectangle& =\sf length \times width\\\\& = 5 \times 1\frac13\\\\ & = \dfrac51\times \dfrac43\\\\& = \dfrac{20}{3}\\\\ & = 6\frac23\: \sf ft^2\end{aligned}$

$\begin{aligned} \sf Area\:of\:irregular\:figure & = \sf area\:of\:triangle+area\:of\:rectangle\\\\ & = 4 + 6\frac23\\\\ & = 10\frac23\: \sf ft^2\end{aligned}$

5 0

2 years ago

Read 2 more answers