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3 years ago

13

There are 132 projects in the science fair. if 8 projects can fit in a row, how many full rows of the projects can be made? how

many projects are in the row that is not full?

2 answers:

VikaD [51]3 years ago

6 0

132/8=16 rows with 4 left over

16 full rows of 8 projects can be made and 4 projects are in the row that is not full.

wel3 years ago

4 0

Divide 132 by 8. You will get the quotient of 16 and the remainder of 4. So each row has 16 projects. The remaining 4 projects cannot be filled completely in a row.

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For ΔABC, which two relationships are true?

Phoenix [80]

The two correct relationships are:

$cos(\theta)=\frac{AB}{AC} = sin(90-\theta)\\\\sin(\theta)=\frac{BC}{AC} = cos(90-\theta)$

<h3>What is a right-angled triangle?</h3>

The triangle shown is a right-angled triangle

Considering < $(90-\theta)^o$

$cos(90-\theta)=\frac{BC}{AC} \\\\sin(90-\theta)=\frac{AB}{AC}\\\\tan(90-\theta)=\frac{AB}{BC}$

Considering < $\theta$

$cos(\theta)=\frac{AB}{AC} \\\\sin\theta)=\frac{BC}{AC}\\\\tan(\theta)=\frac{BC}{AB}$

Comparing < $\theta$ and < $(90-\theta)^o$ , the true statements are:

$cos(\theta)=\frac{AB}{AC} = sin(90-\theta)\\\\sin(\theta)=\frac{BC}{AC} = cos(90-\theta)$

Learn more on trigonometry here: brainly.com/question/20519838

6 0

2 years ago

PLS HELP MEE ASAP 1 MIN LEFT HELPP!!!!!!!!!

artcher [175]

Answer: BC = 5.83

Step-by-step explanation:

Luckily, the triangle is placed on the graph nicely so we can count the legs of the triangle:

AB = 5

AC = 3

BC = ?

To find BC, we can simply use the Pythagorean Theorem:

$a^{2}+b^{2}=c^{2}$

5^2 + 3^2 = c^2

25 + 9 = c^2

34 = c^2

Now square root to find c, or BC.

$\sqrt{34}=\sqrt{c^{2} }$

c = 5.83 (rounded by nearest hundredth)

4 0

3 years ago

sallys cup cake shop sold a total of 63 cupcakes yesterday and 32 of those had sprinkles how many cupcakes were sold without spr

lianna [129]

Answer:

31

Step-by-step explanation:

63-32=31

3 0

3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

if you were asked to load 225 boxes onto a truck and the boxes are crated with each crate containing nine boxes how many crates

jeyben [28]

25

Steps: 225 / 9 = 25 crates

4 0

3 years ago