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3 years ago

12

Sally read 13 books during this month. her brother read 7 books. one fourth out of all books they read were non fiction. how man

y non fiction books did they read?

1 answer:

AveGali [126]3 years ago

7 0

Sally read 3.25 and her brother read 1.75 non fiction

they all read 5 non fiction books all together

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PLEASE HELP I ONLY HAVE 11 MINS LEFT!!!

Novosadov [1.4K]

Answer:

3 and 1/4 / 3/4 = 4 and 1/3

Step-by-step explanation:

The shaded area is 3 1/4 units

The segments are 3/4 units long

The 5th segment is only 1/3 full

So the answer is 4 1/3

3 0

3 years ago

A country has a total biocapacity of 6.21 ha/person, a biocapacity of grazing land of 0.85 ha/person, and a biocapacity of fores

leonid [27]

Answer:

44.8 %

Step-by-step explanation:

Let p = the number of people. Then

Total biocapacity = 6.21p ha

Grazing biocapacity = 0.85p ha

Forest biocapacity = <u>2.53p </u>

Grazing + forest = 2.78p ha

$\text{Percent grazing + forest} = \dfrac{\text{grazing + forest}}{\text{Total}} \times 100 \, \% \\\\= \dfrac{\text{2.78p ha}}{\text{6.21p ha}} \times 100 \, \% = \mathbf{44.8 \,\%}$

6 0

3 years ago

You have a fishing line spool with an end that has an area of 20.0 cm2. how much fishing line do you need to wind around the spo

grandymaker [24]

Answer: 158.53 cm

Step-by-step explanation:

We know that the ends of a fishing line spool are circular in shape.

Given : The area of an end of fishing line spool = $20.0\cm^2$ (1)

Area of a circle = $\pi r^2$ (2)

Circumference of a circle = $2\pi r$ (3)

, where r is radius of the circle.

From (1) and (2), we have

$\pi r^2=20\\\\\Rightarrow\ r^2=\dfrac{20}{\pi}\\\\\Rightarrow\ r=\sqrt{\dfrac{20}{\pi}}$

Circumference of fishing spool = $2\pi r$ (using (3))

$=2\pi \sqrt{\dfrac{20}{\pi}}=2\sqrt{20\pi}=2\times2\sqrt{5\pi}=4\sqrt{5\pi}$

i.e. Fishing spool required to wind around the spool one time $=4\sqrt{5\pi}\ cm$

⇒ Fishing spool required to wind around the spool 10 times $=10\times4\sqrt{5\pi}\ cm=40\sqrt{5\pi}\\\\=40\times\sqrt{5}\times \pi\\\\=40\times2.236\times\sqrt{3.14159}\\\\=89.44\times1.772453\\\\=158.528205473\approx158.53\ cm$

Hence, you need 158.53 cm or about 159 cm of fishing line to wind around the spool 10 times.

7 0

3 years ago

Help me please im in a rush i need help

GREYUIT [131]

$\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star$

Area of shaded region = 34 cm²

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

Area of shaded region = Area of bigger rectangle - Area of smaller rectangle.

Area (big) :

$\qquad❖ \: \sf \:13 \times 6$

$\qquad❖ \: \sf \:78 \: \:c m {}^{2}$

Area (small) :

$\qquad❖ \: \sf \:11 \times 4$

$\qquad❖ \: \sf \:44 \: \: cm {}^{2}$

Area of shaded region :

$\qquad❖ \: \sf \:78 - 44$

$\qquad❖ \: \sf \:34 \: \: cm {}^{2}$

$\qquad \large \sf {Conclusion} :$

Area of shaded region = 34 cm²

8 0

2 years ago

Given: F(x) = 2x^2+ 1, G(x) = 2x - 1, H(x) = x<br> F-2) =

Mnenie [13.5K]

Answer:

Step-by-step explanation:

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3 years ago