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3 years ago

8

Adding Integers -8+19+8?

1 answer:

Eddi Din [679]3 years ago

4 0

-8+19+8 your answer is going to be 19

hope that helps

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A florist uses 10 red roses for every 2 white roses in her bouquets. If the florist uses 10 white roses in an arrangement, how m

ohaa [14]

Answer:

100

Step-by-step explanation:

dont know how to explain it.

7 0

3 years ago

Read 2 more answers

A line passes through the points (2,4) and (5,6) .

Alenkinab [10]

Answer:

Option B and D are correct.

Step-by-step explanation:

Given: A line passes through the points (2,4) and (5,6).

* Case 1:

If a line passes through the points (2, 4) and (5, 6)

Point slope intercept form:

for any two points $(x_1,y_1)$ and $(x_2, y_2)$

then the general form $y -y_1=m(x-x_1)$ for linear equations where m is the slope given by:

$m =\frac{y_2-y_1}{x_2-x_1}$

First calculate slope for the points (2, 4) and (5, 6);

$m = \frac{y_2-y_1}{x_2-x_1} =\frac{6-4}{5-2} = \frac{2}{3}$

then, by point slope intercept form;

$y-4=\frac{2}{3}(x-2)$

* Case 2:

If a line passes through the points (5, 6) and (2, 4)

First calculate slope for the points (5, 6) and (2, 4);

$m = \frac{y_2-y_1}{x_2-x_1} =\frac{4-6}{2-5} = \frac{-2}{-3}= \frac{2}{3}$

then, by point slope intercept form;

$y-6=\frac{2}{3}(x-5)$

Yes, the only equation of line from the given options which describes the given line are;

$y-4=\frac{2}{3}(x-2)$ and $y-6=\frac{2}{3}(x-5)$

8 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.

7 0

3 years ago

D =<br> = (x - x)*+(y - y)<br> +<br> TU = (4,1) (3,-1)<br> Answer:

RideAnS [48]

Answer:

Distance

2.23606797749979

Midpoint

(3.5,0)

Slope

2

x intercept

3.50

y intercept

−7.00

Step-by-step explanation:

6 0

3 years ago

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Use the unique factorization of integers theorem to prove the following statement. If n is any positive integer that is not a pe

anyanavicka [17]

Answer:

no

Step-by-step explanation:

3 0

3 years ago