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3 years ago

14

Subtract (4x+2) from the sum of (6x+1) and (7x+3)

2 answers:

Sphinxa [80]3 years ago

6 0

Answer:11x

Step-by-step explanation:

IRISSAK [1]3 years ago

4 0

Answer:

9x - 6

Step-by-step explanation:

combine like terms = 13x + 4

13x + 4 - (4x + 2)

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Envenenan i don’t know

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3 years ago

Read 2 more answers

Find x.______________a0

velikii [3]

Answer:

$x=4$

Step-by-step explanation:

Before we find x, we need to set up this triangle a little more. We need to find the triangle's altitude before we can solve for x. We will use the heartbeat method to find the altitude.

Let altitude = y; solve for y:

$\frac{2}{y}=\frac{y}{6}$

$y^2=12$

$y=\sqrt{12}$

$y=2\sqrt{3}$

Now that we know the altitude, we can use the Pythagorean Theorem to find the hypotenuse (x).

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

$2^2+(2\sqrt{3})^2=c^2$

$4+12=c^2$

$c^2=16$

$c=4$

Since c and x are the same; c is just the hypotenuse in the Pythagorean Theorem.

$x=4$

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3 years ago

Geometry question help

ololo11 [35]

The answer is b because

7 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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3 years ago

Need help!!! will mark brainliest!!!!!!

irga5000 [103]

Answer:

right 1 unit and down 5 units

Step-by-step explanation:

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3 years ago