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3 years ago

Algebra 1 ) U. Checkpoint: Systems of equations and inequalities LOW

Mathematics

1 answer:

krek1111 [17]3 years ago

7 0

Answer:algebra ok

Step-by-step explanation:

You might be interested in

Solve the following quadratic equation:

Advocard [28]

Answer:

-3.5 or -6

Step-by-step explanation:

2x^2+19x+40=-2

2x^2+19x+40+2=0

2x^2+19x+42=0

2x^2+(12+7)x+42=0

2x^2+12x+7x+42=0

2x(x+6)+7(x+6)=0

(x+6)(2x+7)=0

either,

x+6=0

x=-6

or,

2x+7=0

2x=-7

x=-7/2

x=-3.5

4 0

3 years ago

Rewrite (7x-8)(-1) using the distributive property

stealth61 [152]

The answer would be (-7x+8), the -1 basically means to switch the symbols

4 0

3 years ago

Read 2 more answers

Identify the correct corresponding parts

andrey2020 [161]

Answer:

The correct corresponding part is;

$\overline {CB}$ ≅ $\overline {CD}$

Step-by-step explanation:

The information given symbolically in the diagram are;

ΔCAB is congruent to ΔCED (ΔCAB ≅ ΔCED)

Segment $\overline {CA}$ is congruent to $\overline {CE}$ ( $\overline {CA}$ ≅ $\overline {CE}$ )

Segment $\overline {CB}$ is congruent to $\overline {CD}$ ( $\overline {CB}$ ≅ $\overline {CD}$ )

From which, we have;

∠A ≅ ∠E by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∠B ≅ ∠D by CPCTC

Segment $\overline {AB}$ is congruent to $\overline {DE}$ ( $\overline {AB}$ ≅ $\overline {DE}$ ) by CPCTC

Segment $\overline {AE}$ bisects $\overline {BD}$

Segment $\overline {BD}$ bisects $\overline {AE}$

Therefore, the correct option is $\overline {CB}$ ≅ $\overline {CD}$

3 0

3 years ago

Wyatt is going to a carnival that has games and rides. Each game costs $1. 25 and each ride costs $2. 75. Wyatt spent $20. 25 al

Nonamiya [84]

In the carnival the number of games number of rides Wyatt went on is 3 and 6 respectively.

<h3>What is system of equation?</h3>

A system of equation is the set of equation in which the finite set of equation is present for which the common solution is sought.

The number of unknown objects can be find using the system of solution.

Given information-

The cost of each game is $1.25.

The cost of each ride is $2.75.

Total amount of money spent by the Wyatt is $20.25.

The number of rides he went on is twice the number of games he played.

Suppose the number of games Wyatt played is x and the number of rides he went in the carnival is y.

As, the number of rides he went on is twice the number of games he played. Thus,

$y=2x$

Let the above equation as equation one.

As total amount of money spent by the Wyatt is $20.25 and The cost of each game is $1.25 and the cost of each ride is $2.75. Thus,

$1.25x+2.75y=20.25\\$

Put the value of y in the above equation from the equation one as,

$1.25x+2.75(2x)=20.25\\1.25x+5.5x=20.25\\6.75x=20.25\\x=3$

Thus the number of games he played is 3.

Put this value of x in the equation one as,

$y=2\times3\\y=6$

Thus the number rides Wyatt went on is 6.

Hence, In the carnival the number of games number of rides, Wyatt went on is 3 and 6 respectively.

Learn more about the system of equations here;

brainly.com/question/13729904

5 0

2 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago