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3 years ago

Find the quotient. Show all of your work.

Mathematics

1 answer:

tigry1 [53]3 years ago

3 0

Answer:

0.7

Step-by-step explanation:

Change the divisor 0.525 to a whole number by moving the decimal point 3 places to the right. Then move the decimal point in the dividend the same, 3 places to the right.

We then have the equations:

367.5 ÷ 525 = 0.7

and therefore:

0.3675 ÷ 0.525 = 0.7

Both calculated to 1 decimal places.

I THINK?? HOPE THIS HELPS!!

LIKE IF ITS CORRECT

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4 years ago

Read 2 more answers

The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find t

Galina-37 [17]

Answer:

∠A = ∠C = 60°

∠B = ∠D = 120°

Step-by-step explanation:

In quad. DPBQ, by angle sum property we have

∠PDQ + ∠DPB + ∠B + ∠BQD = 360°

60° + 90° + ∠B + 90° = 360°

∠B = 360° – 240°

Therefore, ∠B = 120°

But ∠B = ∠D = 120° opposite angles of parallelogram

As, AB || CD opposite sides of a parallelogram

∠B + ∠C = 180° sum of adjacent interior angles is 180°

120° + ∠C = 180°

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Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal

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3 years ago

Evaluate 10m + n2/4 when m=5 and n=4

leonid [27]

Answer:

Step-by-step explanation:

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3 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago