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3 years ago

Show examples of perpindicularlines

Mathematics

2 answers:

VladimirAG [237]3 years ago

8 0

A cross, or a + is an example of perpendicular lines. They have to have right angles.

V125BC [204]3 years ago

5 0

Here is an example of perpendicular lines hope this helped and have a great day !!!

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Classify the following polynomial by degree and number of terms: 3x+12

Licemer1 [7]

2 terms and power of 1
Linear binomial

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3 years ago

Does side lengths 4, 8 and 12 form a right triangle?

NeTakaya

Answer:

No it does not form a triangle.

Step-by-step explanation:

If the lengths satisfy the Pythagorean Theorem (a2+b2=c2) then it is a right triangle.

7 0

3 years ago

Read 2 more answers

The formula for the area of a parallelogram is A = bh, where b is the base and h is the height. A parallelogram that is not draw

Ksivusya [100]

The simplified expression for the area of the parallelogram is A = 2x³ - 8x² - 6x + 24

The base of the parallelogram is represented by b

$b = (x - 4)$

The height of the parallelogram is represented by h

$h = (2 {x}^{2} - 6)$

The area of the parallelogram is represented by A

A = bh

A = (x - 4)(2x²- 6)

A = 2x³ - 6x - 8x² + 24

A = 2x³ - 8x² - 6x + 24

The simplified expression for the area of the parallelogram is A = 2x³ - 8x² - 6x + 24

Learn more here: brainly.com/question/25601035

7 0

3 years ago

I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know

azamat

By definition of conditional probability,

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}$

$=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}$

Assuming a standard 6-sided fair die,

if $X_1=4$ , then $X_1+X_2=7$ means $X_2=3$ ; otherwise,
if $X_2=4$ , then $X_1=3$ .

Both outcomes are mutually exclusive with probability $\frac1{36}$ each, hence total probability $\frac2{36}=\frac1{18}$ .

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs $\frac6{36}=\frac16$ of the time.

Then the probability we want is

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13$

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3 years ago

Which algebraic expression matches this scenario?

Vesna [10]

B

If he can type 1/3 as many words as her that would be 1/3j which is the same as j/3.

7 0

3 years ago