5 markers cost $6.55.Which equation would help determine the cost of 4 markers?

Sovle the system of equations using the substitution method (pls show work if you can haha)

lions [1.4K]

Just get photo math and take a pic and it’ll show you how to do it like in elimination substitution and graphing

5 0

3 years ago

(Distribute , then simplify the remaining expression. Final answer must be in standard form) Please help !!

Sergeu [11.5K]

Problem 22
$4x^2(2x^2+x-5) - x(x^3+5x^2-3) + 17$
$4x^2(2x^2)+4x^2(x)+4x^2(-5) - x(x^3)-x(5x^2)-x(-3) + 17$
$8x^4+4x^3-20x^2 - x^4-5x^3+3x + 17$
$7x^4-x^3-20x^2+3x + 17$

So the final simplified answer to problem 22 is
$7x^4-x^3-20x^2+3x + 17$

--------------------------------

Problem 23
$-2x(x^3-6x^2+6)+4x^3-(5x^4+10x)$
$-2x(x^3)-2x(-6x^2)-2x(6)+4x^3-(5x^4)-(10x)$
$-2x^4+12x^3-12x+4x^3-5x^4-10x$
$-7x^4+16x^3-22x$

So problem 23 simplifies to
$-7x^4+16x^3-22x$

---------------------------------------------------------

Problem 24
$4b[2a^2-5(3ab-2b^2)]+29ab^2$
$4b[2a^2-5(3ab)-5(-2b^2)]+29ab^2$
$4b[2a^2-15ab+10b^2]+29ab^2$
$4b[2a^2]+4b[-15ab]+4b[10b^2]+29ab^2$
$8a^2b-60ab^2+40b^3+29ab^2$
$8a^2b-31ab^2+40b^3$

So problem 24 simplifies to
$8a^2b-31ab^2+40b^3$

-------------------------------------------------

Problem 25
$2y(6x-4x^2y)+13x^2y^2-6$
$2y(6x)+2y(-4x^2y)+13x^2y^2-6$
$12xy-8x^2y^2+13x^2y^2-6$
$12xy+5x^2y^2-6$

The final answer for problem 25 is
$12xy+5x^2y^2-6$

8 0

3 years ago

WILL MARK YOU BRAINLIEST, PLEASE HELP!!!

lutik1710 [3]

Given:ABCD is a rhombus.

To prove:DE congruent to BE.

In rombus, we know opposite angle are equal.

so, angle DCB = angle BAD

SINCE, ANGLE DCB= BAD

SO, In triangle DCA

angle DCA=angle DAC

similarly, In triangle ABC

angle BAC=angle BCA

since angle BCD=angle BAD

Therefore, angle DAC =angle CAB

so, opposite sides of equal angle are always equal.

so,sides DC=BC

Now, In triangle DEC and in triangle BEC

1. .DC=BC (from above)............(S)

2ANGLE CED=ANGLE CEB (DC=BC)....(A)

3.CE=CE (common sides)(S)

Therefore,DE is congruent to BE (from S.A.S axiom)

7 0

2 years ago

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

brainly.com/question/24756209

#SPJ4

7 0

1 year ago

Bernita and Derek each plot a number on a number line. The numbers are unique but have the same absolute value. The sum of the a

diamong [38]

The correct answer is B. -75 and 75.

Absolute value means the positive value of a number. Absolute value of -75 is 75. Absolute value of 75 is 75.

75 + 75 = 150.

3 0

3 years ago

Read 2 more answers