All categories

2 years ago

3×4 10×10+10/2-6×4Please help me as soon as possible

Mathematics

1 answer:

Wittaler [7]2 years ago

3 0

You have to replace the variables with the number so your new equation would be 3x16x10+y/2-6x and your answer would be 461

You might be interested in

I NEED HELP IMMEDIATELY PLEASE AND THANK YOU!<br> Solve 7−√159−2=−5

Yuri [45]

Answer:

7-12.6-2=-5

-7.6=-5

7.6-5

2.6

4 0

2 years ago

Two column proofs Beginner's

Darina [25.2K]

LETS prove the 2nd part of the Question

Here concept of right angles is used

We can see BA is perp to BD

and BC is perp to BE

We have to prove angle 1 = angle 3

i will denote angle by a

therefore we need to prove a1 = a3

as BA is perp to BD hence angle between them will be 90 degree

a1 + a2 = 90 degree

Similarly BC is perp to BE

a2 + a3 = 90 degree

as both equations add up gives 90 degree so will equate them

a1 + a2 = a2 + a3

which means a1 = a3

Hence Proved

To learn more about Right Angles:

brainly.com/question/7116550

#SPJ1

3 0

10 months ago

A rectangle has a width of 59 centimeters and a perimeter of 238 centimeters. What is the rectangle's length?

Mademuasel [1]

Answer:

60 cm

Step-by-step explanation:

2x the width and 2x the length gives you the perimeter. This can be represented by the equation:

2w + 2L = P

Use the information given in the word problem to fill in the variables.

2(59) + 2L = 238

118 + 2L = 238

move 118 to the other side of the equation.

2L = 238 - 118

2L = 120

move 2 to the other side of the equation.

L = $\frac{120}{2}$

L = 60

5 0

2 years ago

A long time ago Dulani found an island shaped like a triangle with three straight shores of length 3 km 4 km and 5 km.He said no

VikaD [51]

To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.

This means the exclusion zone would be applied to the base and the height of the triangular space.

You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.

Please see the attached picture to understand this.

You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.

(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².

The exclusion space is 9 km².

5 0

3 years ago

The prices of all college textbooks follow a bell-shaped distribution with a mean of $113 and a standard deviation of $12. Using

Soloha48 [4]

Step-by-step explanation:

In statistics, the empirical rule states that for a normally distributed random variable,

68.27% of the data lies within one standard deviation of the mean.

95.45% of the data lies within two standard deviations of the mean.

99.73% of the data lies within three standard deviations of the mean.

In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

$\Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})$

where the symbol $\Phi$ (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, $\mu$ is the mean and $\sigma$ is the standard deviation of the normal distribution defined as $N(\mu, \ \sigma)$ .

According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution $N(113, \ 12)$ ,

$\Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973$

Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.

5 0

2 years ago