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2 years ago

Add. Write the sum in simplest form. 12 1/3 + 6 2/6 =

Mathematics

2 answers:

gavmur [86]2 years ago

8 0

Answer:

18 2/3

Step-by-step explanation:

12 1/3 + 6 2/6

add the whole numbers

then make the 1/3 into 2/6 bc you need to make like denominators

then add 2/6 and 2/6 which is 4/6

to get simplest form you need to divide by 2 and on top and bottom and get 2/3

(Brainliest?)

Zanzabum2 years ago

7 0

Answer:

18 2/3

Step-by-step explanation:

12 1/3+6 2/6

12 1/3+6 1/3= 18 2/3

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Add. 1/2+(−3/5) can u help me

attashe74 [19]

Exact form: -1/10

decimal form: -0.1

6 0

2 years ago

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A bin contains 64 light bulbs of which 10 are defective. if 5 light bulbs are randomly selected from the bin with replacement, f

Sloan [31]

Total number of bulbs = 64
Number of defective bulbs = 10
Number of good bulbs = 64 - 10 = 54

P(5 good bulbs) = (54/64)⁵ = (27/32)⁵ = 0.428

Answer: 0.428

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2 years ago

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

Find the distance AB

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

Find the distance BC

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

Find the distance AC

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

Find the distance DC

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

Find the perimeter of the triangle

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

Find the area of the triangle

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

4 0

3 years ago

Read 2 more answers

4 3/4 X 4 1/7 what is it please help

Y_Kistochka [10]

first you turn the fractions into inproper fractions by multiplying the whole number by the denomonator and adding the numerator

4×4=16

16+3=19

19/4

4×7=28

28+1=29

29/7

keep the denomonators the same. now that you've converted them into inproper fractions you can multiply them

19/4 × 29/7=551/28

there is no way to simplify this answer so this is the final answer

7 0

2 years ago

If sinA+cosecA=3 find the value of sin2A+cosec2A

Irina18 [472]

Answer:

$\sin 2A + \csc 2A = 2.122$

Step-by-step explanation:

Let $f(A) = \sin A + \csc A$ , we proceed to transform the expression into an equivalent form of sines and cosines by means of the following trigonometrical identity:

$\csc A = \frac{1}{\sin A}$ (1)

$\sin^{2}A +\cos^{2}A = 1$ (2)

Now we perform the operations: $f(A) = 3$

$\sin A + \csc A = 3$

$\sin A + \frac{1}{\sin A} = 3$

$\sin ^{2}A + 1 = 3\cdot \sin A$

$\sin^{2}A -3\cdot \sin A +1 = 0$ (3)

By the quadratic formula, we find the following solutions:

$\sin A_{1} \approx 2.618$ and $\sin A_{2} \approx 0.382$

Since sine is a bounded function between -1 and 1, the only solution that is mathematically reasonable is:

$\sin A \approx 0.382$

By means of inverse trigonometrical function, we get the value associate of the function in sexagesimal degrees:

$A \approx 22.457^{\circ}$

Then, the values of the cosine associated with that angle is:

$\cos A \approx 0.924$

Now, we have that $f(A) = \sin 2A +\csc2A$ , we proceed to transform the expression into an equivalent form with sines and cosines. The following trignometrical identities are used: