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2 years ago

100 points!!! PLEASE ANSWER IT HOW I HAVE IT IN ORDER!!

Mathematics

2 answers:

Anvisha [2.4K]2 years ago

8 0

Answer: Have a good day. I’m sorry you have to go through this, my head already hurts and I am just lōoking at it.

rewona [7]2 years ago

3 0

Answer:

a^2 + b^2 = c^2...a and b are the legs and c is the hypotenuse

90^2 + 120^2 = c^2

8100 + 14400 = c^2

22500 = c^2.....take the sq rt of both sides...this gets rid of the ^2

sq rt 22500 = c

150 = c

so side a = 90 ft, side b = 120 ft, and side c = 150 ft

this is called the pythagorean theorem and it only works on right triangles

Step-by-step explanation:

done sorry if im wrong

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157 29 Marks

AlekseyPX

Complete question :

Put these fraction in order of size smallest to largest : 7/10, 2/3, 4/5, 11/15

Answer:

2/3, 7/10, 11/ 15, 4 /5

Step-by-step explanation:

In other to make solving the question easier, we can convert the fractions to decimal in other to make comparison easier :

7/10 = 0.7

2/3 = 0.667

4 /5 = 0.8

11 /15 = 0.733

Using the placement or how the numbers will appear to the right of a number line ;

0.667, 0.7, 0.733, 0.8

Thus we have :

2/3, 7/10, 11/ 15, 4 /5

3 0

2 years ago

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

Find the midpoint with the endpoints (1,2) (7,2)

den301095 [7]

Answer:

(4,2)

Step-by-step explanation:

The x coordinate of the midpoint is the sum of the x coordinates divided by 2

( 1+7)/2 = 8/2 = 4

The y coordinate of the midpoint is the sum of the y coordinates divided by 2

( 2+2)/2 = 4/2 = 2

The midpoint is (4,2)

3 0

2 years ago

Read 2 more answers

3 less than the quotient of 20 and x

UkoKoshka [18]

It would be (20/x) -3

3 0

2 years ago

Are angles 8 and 15 congruent supplementary or neither

kap26 [50]

Angles 8 and 15 are supplementary.

3 0

2 years ago