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xenn [34]
2 years ago
8

Four angles are formed by the intersection of the diagonals of this quadrilateral. Which statement is NOT true?

Mathematics
2 answers:
DIA [1.3K]2 years ago
8 0

Answer:

no picture

Step-by-step explanation:

HACTEHA [7]2 years ago
5 0

(D): m ∠1 = m ∠2

I got it correct, trust me.

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Graph the equation on the coordinate plane.<br> y=1/2x
azamat
Here is how the graph would look!
Hope this helps :)

8 0
2 years ago
Read 2 more answers
Jim needs to rent a car. A rental company charges $21.00 per day to rent a car and $0.10 for every mile driven.
Ulleksa [173]

Answer:

21d+25\leq 115

Max is 4 days.

Step-by-step explanation:

An inequality is a math statement with more the one solution. The solution has a range of values for which the inequality can be satisfied. It has the same following components as an equation: variables and operations. However, the equal sign is replaced by an inequality sign: ,\leq,\geq.

We start by choosing a variable for an unknown value. Here is it the number of days he rents the car. We choose d.

We know he rents it for $21 pr day or 21d for any number of days. We also know her will drive 250 miles at $0.10 per mile or 0.10(250). We will combine the two into an expression.

21d + 0.10(250)

We know his limit is $115. This means he can rent the car for any number of days until he reaches $115. The charge must be less than or equal to 115.

So we write 21d+0.10(250)\leq 115.

To find the number of days he can rent the car, we solve for d using inverse operations.

21d+0.10(250)\leq 115\\21d+25\leq 115\\21d+25-25\leq 115-25\\21d\leq 90\\\frac{21d}{21} \leq \frac{90}{21}\\d\leq 4.3

This means Jim can rent the car up to 4 days. The 5th day will put the cost over $115.

3 0
3 years ago
How do I solve 2x+3y=7, x-y=1
Fittoniya [83]
<span>2x+3y=7
x-y=1

</span>2x+3y=7
x=1+y

2*(1+y)+3y=7
x=1+y

2+2y+3y=7
x=1+y

5y=5       |(:5)
x=1+y

y=1
x=2

CHECK:
2x+3y=7
x-y=1

2*2 + 3*1 = 7     

x-y = 1
2-1 = 1

5 0
3 years ago
Read 2 more answers
For each expression, select all equivalent expressions from the list.
ratelena [41]

\qquad\qquad\huge\underline{{\sf Answer}}

Let's find the equivalent expressions ~

#1. PROBLEM

\qquad \tt \dashrightarrow \:11x - 4x + 10x

\qquad \tt \dashrightarrow \:21x - 4x

\qquad \tt \dashrightarrow \:17x

#2. PROBLEM

\qquad \tt \dashrightarrow \:2(1 + 8y)

\qquad \tt \dashrightarrow \:(2 \times 1) + (2 \times 8y)

\qquad \tt \dashrightarrow \:2 + 16y

8 0
2 years ago
Math help guys how wouls i work this out
Andrei [34K]

8 > 7 + \frac{x}{6}   Subtract 7 from both sides

1 > \frac{x}{6}   Multiply both sides by 6

6 > x   Flip it around so it's easier to read

x < 6

You can graph your answer by drawing an open circle at the 6 and coloring the line to the left. The circle should be open, because x is <em>less than 6</em>, not less than or equal to. You would color to the left to show that x can be anything less than 6.

4 0
3 years ago
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