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3 years ago

{[(8-3)x2]+[5x6)-5]} divided by five

2 answers:

6 0

{[(8 - 3) * 2] + [(5 * 6) - 5]} / 5 =
= {[2 * 8 + 2 * (-3)] + 30 - 5} / 5 =
= [(16 - 6) + 25] / 5 =
= (10 + 25) / 5 =
= 35 / 5 =
= 7

Brackets make it look complex but it's not that bad at all :)

nignag [31]3 years ago

3 0

Answer:

The answer is 7.

Step-by-step explanation:

First we will solve this as per pemdas.

${[(8-3)\times2]+[(5\times6)-5]}$

= ${[5\times2]+[(5\times6)-5]}$

=> ${[10]+[(30)-5]}$

=> ${[10]+[25]}$

= 35

Now 35 divided by 5 = 7

$\frac{35}{5}=7$

Therefore, the answer is 7.

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Find the length x to the nearest whole number. A right triangle has a vertical leg of length x units, a base leg with a length o

wel

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value of x is $x = 274 \ unites$

Step-by-step explanation:

From the question we are told that

The length of the vertical length is $x$

The length of the base leg is L = 330 + d units

The length of the bisecting line segment is h

The base angle is $\theta = 28^o$

The angle between d and line segment is $\theta_1 = 56^o$

For the first angle

$Tan \theta_1 = \frac{x}{d}$

=> $d = \frac{x}{Tan \theta _1}$

For the whole big triangle

$Tan \theta = \frac{x}{330 + d}$

=> $d = \frac{x}{Tan \theta } -330$

So equating the both d

$\frac{x}{Tan \theta _1} = \frac{x}{Tan \theta } -330$

Substitute values

$\frac{x}{Tan (56)} = \frac{x}{Tan (28) } -330$

$0.6745 x = 1.880x -330$

$1.20549 x = 330$

$x = 274 \ unites$

5 0

3 years ago

What is the sum of the first 70 consecutive odd numbers? Explain.

expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1$

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

$\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)$

The second sum is simply the sum of N ones:

$\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N$

So, the final result is

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2$

which ends the proof.

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3 years ago

Chloe wants to cut a 37 in. piece of ribbon into three pieces. The second piece is to be 2 in. longer than the first, and the th

valentinak56 [21]

1st piece = x

2nd piece = x+2

3rd piece = x+2+3 = x+5

37 = x +x+2 + x+5 =

37 = 3x+7

30=3x

x = 30/3 = 10

1st piece = 10 inches

2nd piece = 10+2 = 12 inches

3rd piece = 10+5 = 15 inches

7 0

3 years ago

Look at the street map. What street intersects with Geary Street?

Vinvika [58]

A. Union Street

Hope This Helps!

8 0

2 years ago

Read 2 more answers

Find the area of the figure.

jasenka [17]

Divide into shapes that you can find the area of.....

you also have to cross out the 5 and the 4s because you will not need them

2×2= 4 sq. cm
2×2= 4 sq. cm
9×2 = 18 sq. cm

18 + 4 + 4 = 26 sq. cm

so the area is 26 sq. cm

6 0

3 years ago