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Masteriza [31]
2 years ago
12

Jake’s car has 33 gallons of gas on the tank at the beginning of his road trip Jake uses 8 gallons of gas every hour he drives.

( look at the picture for answer choices ) plsss help me!!!

Mathematics
2 answers:
34kurt2 years ago
5 0
I think y=33x-8 because if you use 8 gallons every hour you won’t be gaining your losing gas so it’s not adding or multiplying so the next best thing is A!
tamaranim1 [39]2 years ago
5 0

Answer:

Anser is A

Step-by-step explanation:

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I have SAT's tomorrow and need help with math...
Brrunno [24]

Answer:

3w + 52 ≥ 90

Step-by-step explanation:

If she volunteers 3 hours per week for w weeks, then the number of hours from tutoring is 3w.  Add the 52 hours from her summer volunteering, and her total hours is 3w + 52.  This needs to be at least 90 hours for her to graduate, so:

3w + 52 ≥ 90

3 0
3 years ago
What Ln(z) is answer????!!
dimaraw [331]

Write <em>z</em> in exponential form:

z=1-i=\sqrt2 e^{-i\frac\pi4}

Then taking the logarithm, we get

\mathrm{Ln}(z)=\ln(\sqrt2) + \ln e^{-i\frac\pi4} = \boxed{\ln(\sqrt2)-\dfrac\pi4i}

so a is the correct answer.

3 0
3 years ago
If x = (√2 + 1)^-1/3 then the value of x^3 + 1/x^3 is​
Shtirlitz [24]

Step-by-step explanation:

<u>Given</u><u>:</u> x = {√(2) + 1}^(-1/3)

<u>Asked</u><u>:</u> x³+(1/x³) = ?

<u>Solution</u><u>:</u>

We have, x = {√(2) + 1}^(-1/3)

⇛x = [1/{√(2) + 1}^(1/3)]

[since, (a⁻ⁿ = 1/aⁿ)]

Cubing on both sides, then

⇛(x)³ = [1{/√(2) + 1}^(1/3)]³

⇛(x)³ = [(1)³/{√(2) + 1}^(1/3 *3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(1*3/3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(3/3)]

⇛(x * x * x) = [(1*1*1)/{√(2) + 1)^1]

⇛x³ = [1/{√(2) + 1}]

Here, we see that on RHS, the denominator is √(2)+1. We know that the rationalising factor of √(a)+b = √(a)-b. Therefore, the rationalising factor of √(2)+1 = √(2) - 1. On rationalising the denominator them

⇛x³ = [1/{√(2) + 1}] * [{√(2) - 1}/{√(2) - 1}]

⇛x³ = [1{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Multiply the numerator with number outside of the bracket with numbers on the bracket.

⇛x³ = [{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Now, Comparing the denominator on RHS with (a+b)(a-b), we get

  • a = √2
  • b = 1

Using identity (a+b)(a-b) = a² - b², we get

⇛x³ = [{√(2) - 1}/{√(2)² - (1)²}]

⇛x³ = [{√(2) - 1}/{√(2*2) - (1*1)}]

⇛x³ = [{√(2) - 1}/(2-1)]

⇛x³ = [{√(2) - 1}/1]

Therefore, x³ = √(2) - 1 → → →Eqn(1)

Now, 1/x³ = [1/{√(2) - 1]

⇛1/x³ = [1/{√(2) - 1] * [{√(2) + 1}/{√(2) + 1}]

⇛1/x³ = [1{√(2) + 1}/{√(2) - 1}{√(2) + 1}]

⇛1/x³ = {√(2) + 1}/[{√(2) - 1}{√(2) + 1}]

⇛1/x³ = [{√(2) + 1}/{√(2)² - (1)²}]

⇛1/x³ = [{√(2) + 1}/{√(2*2) - (1*1)}]

⇛1/x³ = [{√(2) + 1}/(2-1)]

⇛1/x³ = [{√(2) + 1}/1]

Therefore, 1/x³ = √(2) + 1 → → →Eqn(2)

On adding equation (1) and equation (2), we get

x³ + (1/x³) = √(2) -1 + √(2) + 1

Cancel out -1 and 1 on RHS.

⇛x³ + (1/x³) = √(2) + √(2)

⇛x³ + (1/x³) = 2

Therefore, x³ + (1/x³) = 2

<u>Answer</u><u>:</u> Hence, the required value of x³ + (1/x³) is 2.

Please let me know if you have any other questions.

3 0
2 years ago
What does it mean when it says a shape has reflection symmetry?
brilliants [131]
It means that if you fold it in half the two halves will be the exact same
6 0
3 years ago
2 Here are two equations:
MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

<u>i (3,4) </u>

Putting in Equation 1:

6(3) + 4(4) = 34\\18+16=34\\34=34\\

Putting in Equation 2:

5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15

<u>ii. (4,2.5)</u>

Putting in Equation 1:

6(4) + 4(2.5) = 34\\24+10=34\\34=34\\

Putting in Equation 2:

5(4) - 2(2.5) = 15\\20-5 = 15\\15=15

<u>ill. (5,5)</u>

6(5) + 4(5) = 34\\30+20=34\\50\neq 34

Putting in Equation 2:

5(5) - 2(5) = 15\\25-10 = 15\\15=15

<u>iv. (3,2)</u>

6(3) + 4(2) = 34\\18+8=34\\26\neq 34

Putting in Equation 2:

5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

  • brainly.com/question/10534381
  • brainly.com/question/10538663

#LearnwithBrainly

8 0
3 years ago
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