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Ymorist [56]
3 years ago
12

How would I factor x2 + 13x + 22

Mathematics
1 answer:
lozanna [386]3 years ago
6 0
I hope this helps you

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True or false 2(a + 6) is equivalent to 2a + 13?
katen-ka-za [31]

Answer:

false

Step-by-step explanation:

5 0
3 years ago
Need to know angle 1, angle 2, angle 3
wolverine [178]

Answer: Step-by-step explanation:

angle 1=41°

angle 2=angle 1=41°

angle 3=180-2×41=98

4 0
3 years ago
30 POINTS! PLEASE HELP
Kryger [21]

Answer:

y = -4x+4

Step-by-step explanation:

Pick two points on the line

(0,4) and (1,0)

We can find the slope

m = (y2-y1)/(x2-x1)

   = (0-4)/(1-0)

   = -4/1 =-4

We know the y intercept is 4

The slope intercept form of the equation is y = mx+b  where m is the slope and b is the y intercept

y = -4x+4

8 0
3 years ago
Read 2 more answers
10.1 and 10.5 witch one is greater
OLga [1]

Answer:

10.5

Step-by-step explanation:

Because the digit at tenths place in 10.5 is greater than that of in 10.1

5 > 1

3 0
3 years ago
Read 2 more answers
the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger
Reil [10]
<h2>Answer:</h2>

x\geq 3 \ and \ x+1 \geq 4

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

So:

1. The sum of 2 consecutive integers can be written as:

v+(v+1)

2. Nine times the smaller and 5 times the larger can be written as:

9v-5(v+1)

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4

The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

6 0
3 years ago
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