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2 years ago

8

PLEASSS HELPPPP

1 answer:

Lera25 [3.4K]2 years ago

6 0

Answer:

-11 + 4

Step-by-step explanation:

When you subtract a negative the sign will switch to plus, so it this case you would switch the - (-4) to a + 4 but you don't change the -11 at the front.

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What composition of rigid motions maps PQR to XZY?

Lapatulllka [165]

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Step-by-step explanation:

translated 8 units to the right and 2 units up then flipped

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3 years ago

What value of X makes equation below true 2x + 9 = 4x - 1

ale4655 [162]

Answer:

5

Step-by-step explanation:

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2 years ago

Read 2 more answers

Michelle has x dollars. Tony has nine fewer dollars than Michelle. Drag and drop the expressions into the boxes to write an expr

Radda [10]

your answer is x+9+x

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2 years ago

Read 2 more answers

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.

7 0

2 years ago

CAN SOMEONE PLEASE HELP ME WITH MY MATH ASAP PLEASE!!!!

Ivanshal [37]

Answer:

6

Step-by-step explanation:

Hello,

The two triangles are <u>similar</u> and x is <u>positive</u>, <u>different from 0</u>, so we can write.

$\dfrac{x}{9}=\dfrac{12}{x}\\\\x^2=12\times 9\\\\ x = \sqrt{4\times 3 \times 9}=2\times 3 \times \sqrt{3}\\\\ \boxed{\sf \bf x=6\sqrt{3} }$

Thanks

8 0

2 years ago