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Georgia [21]
3 years ago
5

José correctly answered 80% of the questions on a language arts quiz. If he answered 16 questions correctly, how many questions

were on the language arts quiz?
Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0
There were 20 questions on the quiz
nalin [4]3 years ago
6 0
Jose had 20 questions on his test
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A rectangular sheet of paper is 12 1/2 cm long and 10 2/3 cm wide.Find it's perimeter​
ra1l [238]

Answer:

Given

length of rectangular sheet of paper is 12 (1/2) i.e. (25/2)

Breadth of rectangular sheet of paper is 10 (2/3) i.e. (32/3)

But we know that perimeter of rectangle = 2 (length + breadth)

Perimeter of rectangular sheet = 2 [(25/2) + (32/3)]

LCM of 2 and 3 is 6,

by taking this and simplifying we get

Perimeter = 2[(25 × 3)/6 + (32 × 2)/6]

= 2[(75/6) + (64/6)]

= 2(139/6) = (139/3)

= 46 (1/3) cm

3 0
3 years ago
Read 2 more answers
You are dividing 30 tomato plants into equal rows for planting. how many possibilities are there?​
Ierofanga [76]

Answer:

6,5 5,6 3,10 10,3 1,30 30,1 2,15 15,2 = 8 possibilities

Step-by-step explanation:

8 0
2 years ago
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Joshua currently does a total of 8 pushups each day. He plans to increase the number of pushups he does each day by 2 pushups un
alekssr [168]
He currently does 8, and is increasing by 2, so our equation will be

2x+8=30

Subtract the 8 over

2x=22

Divide by 2 on both sides

X=11

It will take 11 days
6 0
3 years ago
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A rectangle has a perimeter of 48 feet and a length of 14 feet. Which equation can you solve to find the width?
xeze [42]
48 = 14*2 + 2n
2n + 28 = 48 
Ans. A
4 0
3 years ago
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Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.
ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region \mathcal T by computing the double integral of the density function over \mathcal T:

\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy

The boundary of \mathcal T is determined by a set of lines in the x,y plane. One way to describe the region \mathcal T is by the set of points,

\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}

So the mass is

\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx

=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx

=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx

=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52

6 0
3 years ago
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