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3 years ago

Solve this problem 3 Times 7

Mathematics

2 answers:

77julia77 [94]3 years ago

7 0

Answer:

3x7=21

Step-by-step explanation:

7+7+7=21 or 3+3+3+3+3+3+3=21

Brainliest

34kurt3 years ago

4 0

Answer:

3 times 7 is 21

Step-by-step explanation:

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Sequence of numbers 1.5,2.25,3.0,3.75 in a recursive formula?

stellarik [79]

Answer:

f(n + 1) = f(n) + 0.75

Step-by-step explanation:

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3 years ago

Solve for z. 3 + 9z = –7 + 10z z =

stiv31 [10]

The answer should be z = 10

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3 years ago

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2, 2, 3 can be lengths of the sides of a triangle. true or false

Vinil7 [7]

Answer:

Yes

Step-by-step explanation:

The sum of any 2 sides of a triangle are greater than the 3rd side.

2,2,3 satisfies this condition.

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4 years ago

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Complete the assignment on a separate sheet of paper Please attach pictures of your work.

Irina18 [472]

Answer:

TO FIND :-

Length of all missing sides.

FORMULAES TO KNOW BEFORE SOLVING :-

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

SOLUTION :-

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

$=> x = \frac{7}{0.27563....} = 25.39568.....$ ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

5 0

3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

5 0

3 years ago

Solve this problem 3 Times￼ 7

Solve this problem 3 Times 7