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4 years ago

To divide two fractions, first rewrite the problem as the dividend times the ______ of the divisor.

Mathematics

2 answers:

gogolik [260]4 years ago

7 0

Answer:

To divide two fractions, first rewrite the problem as the dividend times the _inverse_ of the divisor.

Step-by-step explanation:

Suppose we have the following division of fractions:

$\frac{\frac{b}{c}}{\frac{h}{p}}$

The dividend is the expression that appears in the numerator. In this case, the dividend is: $\frac{b}{c}$

The divisor is: $\frac{h}{p}$

We can write this expression as:

$\frac{b}{c}*\frac{1}{\frac{h}{p}}$

But $\frac{1}{\frac{h}{p}}=\frac{p}{h}$

and $\frac{p}{h}$ is the inverse of $\frac{h}{p}$

Therefore

$\frac{\frac{b}{c}}{\frac{h}{p}}=\frac{b}{c}*\frac{p}{h}$

Then to divide two fractions, first rewrite the problem as the dividend times the _inverse_ of the divisor.

Alika [10]4 years ago

6 0

Answer:

reciprocal

Step-by-step explanation:

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RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that

schepotkina [342]

Answer:

Question 11:

$\angle DAC = 53^\circ$

$\angle AED = 90^\circ$

$\angle ADC = 74$

$DB = 16$

$AE = 6.03$

$AC = 12.06$

Question 12:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

Question 13:

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

Question 11

Given

$\angle BAC = 53^\circ$

$DE = 8$

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): $\angle DAC$

Diagonal CA divides $\angle DAB$ into 2 equal angles

i.e

$\angle DAC = \angle BAC$

So:

$\angle DAC = 53^\circ$

Solving (b): $\angle AED$

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

$\angle AED = 90^\circ$

Solving (c): $\angle ADC$

First, we calculate $\angle ADE$ , considering $\triangle ADE$ :

$\angle ADE + \angle AED + \angle DAC = 180$

$\angle ADE + 90 + 53 = 180$

$\angle ADE + 143 = 180$

$\angle ADE = -143 + 180$

$\angle ADE = 37$

To calculate $\angle ADC$ , we have:

$\angle ADC = 2*\angle ADE$

$\angle ADC = 2* 37$

$\angle ADC = 74$

Solving (d): $DB$

From the rhombus

$DB = DE +EB$

Where

$DE =EB$

So:

$DB = 8 + 8$

$DB = 16$

Solving (e): $AE$

To do this we consider $\triangle ADE$

Using the tan formula

$tan(\angle ADE) = \frac{AE}{DE}$

$\angle ADE = 37$ and $DE = 8$

So:

$\tan(37) = \frac{AE}{8}$

$AE = 8 * \tan(37)$

$AE = 6.03$

Solving (f): $AC$

This is calculated as:

$AC = AE + EC$

Where

$AE = EC$

$AC = 6.03 +6.03$

$AC = 12.06$

Question 12: Isosceles Triangle

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

Question 13:

AC and BD are perpendicular lines, and they are diagonals

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3 years ago

Rob's school is selling tickets to annual dance competition. On the first day of ticket sales the school sold 10 senior citizen

Inga [223]

Answer:

senior tickets are 12

student tickets are 3

Step-by-step explanation:

s= senior

t = student

10s+5t =135

14s + 3t=177

10s+5t =135

divide by 5

2s+t = 27

multiply by -3

-3(2s+t) = -3*27

-6s -3t =-81

add this to 14s+3t=177

-6s -3t =-81

14s + 3t=177

---------------------

8s = 96

divide by 8

s = 12

senior tickets are 8 dollars

2s+t = 27

2(12) +t = 27

24 +t =27

subtract 24

24+t-24 = 27 -24

t =3

student tickets are 3

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4 years ago

In ΔTUV, the measure of ∠V=90°, the measure of ∠T=55°, and VT = 56 feet. Find the length of TU to the nearest tenth of a foot.

HACTEHA [7]

Answer:

97.6

Step-by-step explanation:

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3 years ago

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4. Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean

Karo-lina-s [1.5K]

Answer:

Yes, Tom must be admitted to this university.

Step-by-step explanation:

We are given that the scores on national test are normally distributed with a mean of 500 and a standard deviation of 100.

Also, we are provided with the condition that Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test.

Let, X = score in national test, so X ~ N( $\mu=500 , \sigma^{2} = 100^{2}$ )

The standard normal z distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Now, z score of probability that tom scores 585 is;

Z = $\frac{585-500}{100}$ = 0.85

Now, proportion of students scoring below 85% marks is given by;

P(Z < 0.85) = 0.80234

This shows that Tom scored 80.23% of the students who took test while he just have to score more than 70%.

So, it means that Tom must be admitted to this university.

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4 years ago

the total surface area of a cube 1350 in^2. what is the length of each side cube of the cube it is __in

Svet_ta [14]

Each side is 225 in ^3

5 0

3 years ago

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