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3 years ago

HELP ME ASAP NOW PLEASE!

Mathematics

1 answer:

Aleksandr [31]3 years ago

8 0

1760
13+25+10/150=.32
.32x5500=1760

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1.5d + 3.25 = 1 + 2.25d ---What are the steps to find x?

yuradex [85]

Do u mean d?

1.5d +3.25 = 1 +2.25d
3.25 = 1 + .75d -1.5d on both sides
2.25 = .75d subtraction 1 on both sides
3 = d divide both sides by .75

3 0

4 years ago

The formula for distance traveled d is d=rt, where r is the average speed of the object and t is time traveled. If an object tra

Setler79 [48]

Answer:

22.5

Step-by-step explanation:

d= 75(.30)

5 0

3 years ago

Ashley is making bracelets to sell. She wants to make at least $450. Each bracelet costs $18 to make. She is selling them for $3

Anvisha [2.4K]

Answer:

144 bracelets

Step-by-step explanation:

Ashley is making bracelets to sell. She wants to make at least $450. Each bracelet costs $18 to make. She is selling them for $30 each. What is the minimum number she can sell to reach her goal

Let us represent the minimum number of bracelets she can sell= x

Hence, our Equation is

$18 + $30 × x ≤ $450

$18 + 3x ≤ 450

3x ≤ $450 - $18

3x ≤ $432

x ≤ $432/3

x ≤ 144

The minimum number of bracket = 144 bracelets

8 0

3 years ago

A, B, and C are points on the circumference of a circle, centre 0. AOB is a diameter of the circle. Prove that angle ACB is 90°

ki77a [65]

Answer:

Kindly refer to explanation.

Step-by-step explanation:

Given that:

A,B and C are 3 points on circumference of circle with centre O.

AOB is diameter.

Kindly refer to the attached image.

To prove:

$\angle ACB = 90^\circ$

Solution:

Let $\angle A = x$ and $\angle B = y$ .

<em>Construction</em>: Join point A with centre O.

Considering $\triangle AOC$ :

Side AO = OC because both are the radius.

Angles opposite to equal sides in a triangle are equal.

Therefore, $\angle ACO=\angle A=x$

Considering $\triangle BOC$ :

Side BO = OC because both are the radius.

Angles opposite to equal sides in a triangle are equal.

Therefore, $\angle BCO=\angle B=y$

$\angle ACB = \angle ACO + \angle BCO = x+y$ ...... (1)

Now, in $\triangle ABC:$

Using angle sum property of triangle:

$\angle A + \angle B + \angle ACB =180^\circ\\\Rightarrow x+y+x+y=180^\circ\\\Rightarrow 2(x+y)=180^\circ\\\Rightarrow x+y=90^\circ$

By equation (1):

$\angle ACB = 90^\circ$

Hence proved.

5 0

3 years ago

PLEASE HELP DONE BY TODAY THANK YOU 13-15

ivolga24 [154]

I got the same worksheet lemme look for it

8 0

4 years ago