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Answer:

70-25=45

45+25=70

Step-by-step explanation:

Answer:

[tex]\dfrac{w}{z}=\dfrac{2}{7}[/tex]

Step-by-step explanation:

Given that,

[tex]w=\dfrac{6}{7}[/tex]

z = 3

We need to find [tex]\dfrac{w}{z}[/tex].

Put w = 6/7 and z = 3 in the w/z

[tex]\dfrac{w}{z}=\dfrac{\dfrac{6}{7}}{3}\\\\=\dfrac{6}{7}\cdot\dfrac{1}{3}\\\\=\dfrac{2}{7}[/tex]

So, [tex]\dfrac{w}{z}=\dfrac{2}{7}[/tex]. Hence, the correct answer is 2/7.

Answer:

-3/4

Step-by-step explanation:

-1 1/4 = -5/4

1/2 = 2/4

-5 + 2 = -3

-3/4

Answer: The answer is -3/4

Step-by-step explanation: covert the mixed numbers to improper fractions, then find the LCD and combine.

Step-by-step explanation:

25x² + 100 = 0

25(x² + 4) = 0

x² + 4 = 0

x² = -4

Since x² >= 0 for all real values of x, there is no real solution for x. However there are complex solutions.

=> x = ±√-4

=> x = 2i or x = -2i.

Answer:

daaaaaaaaaa

Step-by-step explanation:

ssdddsss

Answer: 829 3/4 or 829 75/100

Step-by-step explanation:

Answer:

a)  f(5) = 24

b) The inverse of given function

              f⁻¹ ( x ) = √x+1

c)    f⁻¹ ( 8 ) = √9 =3

Step-by-step explanation:

Explanation:-

Given  f(x) = x² - 1

a)

      f(5) = 5² -1 = 25-1 =24

b)      

put y =  f(x) = x² - 1

   ⇒   y =  x² - 1

 ⇒     x² = y + 1

⇒     x = √y+1

⇒ f⁻¹ ( y ) = √y+1      ( ∵ f⁻¹ ( y) =x)

The inverse of given function

              f⁻¹ ( x ) = √x+1

c)   put x=8

             f⁻¹ ( 8 ) = √8+1 = √9 =3

Answer:

Equation of straight line that passes through (-4, 2) and has slope 9/2 is

[tex]y-2=\frac{9}{2} (x+4)[/tex]

Step-by-step explanation:

Equation of straight line in point slope form is given as

[tex]y-y_1=m(x-x_1)[/tex] ---------(2)

Here

[tex]m = \frac {9}{2}[/tex]   and  [tex](x_1, y_1) = (-4, 2)[/tex]

Substituting values in equation (1)

[tex]y-2=\frac{9}{2} (x+4)[/tex]

Answer:

Step-by-step explanation:

1/6 x 1/5 + 1/30

= 1/30 + 1/30

= 2/30

=1/15

Answer:

-3

Step-by-step explanation:

Answer:

B -3

Step-by-step explanation:

x- coordinate of a point is called abscissa and y-coordinate is called ordinate.

x-coordinate of point (-3, 5) is -3

Answer: 22

Step-by-step explanation:

Solve for x

   1/2x + 7 = 18

Combine 1/2 and x.

  x/2+ 7 = 18

Move all terms not containing x to the right side of the equation.

Subtract 7 from both sides of the equation.

  x/2= 18 − 7

  x/2 = 11

Multiply both sides of the equation by 2.

  2 ⋅ x/2 = 2 ⋅ 11

Simplify both sides of the equation.

Cancel the common factor of 2.

  x = 2 ⋅ 11

Multiply 2 by 11.

  x = 22

Answer:

here you go...........

Answer:

%18 tip

Step-by-step explanation:

The dinner cost was $29 before tip

after tip it was $32.22

34.22-29= $5.22  she tipped

$29 * %x = $5.22

After solving this equation, the x is %18

Answer:

angle JKL is 21

Step-by-step explanation:

the angle of the two triangles are the same, so (2x + 1) = (3x - 9)

You would then find the x, which equals to 10.

Then replace x with ten with the equation.

Answer:

yeah I would say deal #1 is better

Answer:

29.8÷3=9.93333333333

Step-by-step explanation:

*note* the 3 is a repeating number

Answer:

b

Step-by-step explanation:

Answer:

No significant difference between mean and Expected value

Step-by-step explanation:

Hypothesis

H0 : u = 77.5

H1: u is not equal to 77.5

Alpha = 0.05

Mean = Σxi/n

= 77.09+75.37+72.42+76.84+77.84, +76.69+78.03+74.96+77.54+76.09+81.12+75.75/12

= 919.74/12

= 76.645

We get the variance s² = xi² -n(barx)²/n-1

= 77.09²+75.37²...75.75²-12(76.645)²/11

When we solve this out

We get 4.3486818182

T test = barx - u/(s/√n)

= (76.645-77.5)/√4.3486818182/12

= -1.420293

T critical = Tn-1, alpha/2

= 2.200985

The test statistic is less than 2.201 so we accept the null hypothesis at 0.05 level of significance. And then conclude that there is No significant difference between mean and Expected value.

Let the two number is a and b

so,

product =ab=20

sum of square=[tex]\bold{a^2+b^2=41 }[/tex]

Then,

[tex]\bold{(a+b)^2=a^2+b^2+2ab }[/tex]

[tex]\bold{ (a+b)^2=41+2×40 }[/tex]

[tex]\bold{ (a+b)^2=81 }[/tex]

[tex]\bold{a+b=\sqrt{81} }[/tex]

[tex]\bold{a+b=9 }[/tex]•••••••••(equation I)

Now,

[tex]\bold{(a-b)^2=a^2+b^2-4ab }[/tex]

[tex]\bold{ (a-b)^2=41-4×20 }[/tex]

[tex]\bold{(a-b)^2=41-40 }[/tex]

[tex]\bold{a-b=\sqrt{1} }[/tex]

[tex]\bold{a-b=1 }[/tex]••••••••(equation II)

Now,combine the equation I and equation II

we,get

[tex]\bold{a+b+a-b=9+1 }[/tex]

[tex]\bold{a+\cancel{b}+a\cancel{-b}=10 }[/tex]

[tex]\bold{ 2a=10 }[/tex]

[tex]\bold{a=\dfrac{10}{2} }[/tex]

[tex]\blue{\boxed{ a=5 }}[/tex]

Then,

put the value of a in equation II.

we get that,

[tex]\bold{5-b=1 }[/tex]

[tex]\bold{b+1=5 }[/tex]

[tex]\bold{b=5-1 }[/tex]

[tex]\bold{\boxed{\blue{b=4}} }[/tex]

so,

The two number is 5 and 4.

Answer:

17 units

Step-by-step explanation:

Use distance formula, [tex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex], to find the distance between (-9, 5) and (8, 5):

Let,

[tex] (x_1, y_1) = (-9, 5) [/tex]

[tex] (x_2, y_2) = (8, 5) [/tex]

Plug in the values into the distance formula:

[tex] d = \sqrt{(8 - (-9))^2 + (5 - 5)^2} [/tex]

[tex] d = \sqrt{(17)^2 + (0)^2} [/tex]

[tex] d = \sqrt{(289 + 0)} [/tex]

[tex] d = \sqrt{289} [/tex]

[tex] d = 17 [/tex]

Answer:

The chart has the steepestt slope

Step-by-step explanation:

The graph has a slope of -3 since it is falling

ANd the table gets you a slope of 1

1>-3

So the table/chart has the steeper slop

Answer:

0.55 > 0.525

Step-by-step explanation:

Answer:

Step-by-step explanation:

0.55=0.550

0.550>0.525

so 0.55>0.525

Step-by-step explanation:

[tex]x ^{1} = - 5 \\ y ^{1} = - 2 \\ x {}^{2} = - 2 \\ y {}^{2} = 1 \\ then \: you \: solve \: with \: that \: formula \: you \: know[/tex]

Answer:

The slope intercept form is y=1/6x+1

Step-by-step explanation:

Answer:

first point goes on (0,1)

second point goes on (6, 1)

hope that helps :)

Step-by-step explanation:

(x is first, y is second)

Answer:

B and E are indeed correct

Answer:

30

Step-by-step explanation:

I could be wrong

The usual speed of Eric's mother is 4 miles per hour .

Eric´s mother drives to work in 20 min when driving at her usual speed.

Let the usual speed of Eric's mother be "u"  miles per hour (mph)

When the traffic is bad she drives 10 miles per hour slower.

The speed of Eric's mother when the traffic is bad = u-10 mph

The time taken by Eric's mother when the traffic is bad = 30 min

[tex]\rm Displacement = \dfrac{Velocity }{Time }[/tex]

The considering the displacement  in both  the cases is same hence we can

say that

[tex]\rm \dfrac{u}{20/60} = \dfrac{-(u-10)}{30/60} \\\rm 3u = -2u +20 \\5u = 20 \\u = 4[/tex]

So the usual speed of Eric's mother is 4 miles per hour .

For more information please refer to the link given below

https://brainly.com/question/23294846

Answer:

Proved all parts below.

Step-by-step explanation:

As given ,

|x| = [tex]\left \{ {{x , x\geq 0} \atop {-x, x< 0}} \right.[/tex]

To prove- a) For any real number x , | x | ≥ 0 . Moreover, | x | = 0 ⇒ x = 0

                 b) For any two real numbers x and y , | x | ⋅ | y | = | x y | .

                 c) For any two real numbers x and y , | x + y | ≤ | x | + | y | .

Proof -

a)

As given x is a real number

Also , by definition of absolute value of x , we get

| x | ≥ 0

Now,

if |x| = 0

⇒ x = 0 and -x = 0

⇒ x = 0 and x = 0

⇒ x = 0

∴ we get

| x | = 0 ⇒  x = 0

Hence proved.

b)

To prove - | x | ⋅ | y | = | x y |

As we have,

|x| = [tex]\left \{ {{x , x\geq 0} \atop {-x, x< 0}} \right.[/tex]

|y| = [tex]\left \{ {{y , y\geq 0} \atop {-y, y< 0}} \right.[/tex]

|xy| = [tex]\left \{ {{xy , x,y > 0 and x,y < 0} \atop {-xy, x > 0, y< 0 and x <0 , y > 0}} \right.[/tex]

We have 4 cases : i)  when x > 0 , y > 0

                              ii) when x > 0 , y < 0

                              iii) when x < 0, y > 0

                              iv) when x < 0, y < 0

For Case I - when x > 0 , y > 0

⇒ |x| = x, |y| = y

⇒|x|.|y| = xy

For Case Ii - when x > 0 , y < 0

⇒ |x| = x, |y| = -y

⇒|x|.|y| = -xy

For Case Iii - when x < 0 , y > 0

⇒ |x| = -x, |y| = y

⇒|x|.|y| = -xy

For Case IV - when x < 0 , y < 0

⇒ |x| = -x, |y| = -y

⇒|x|.|y| = (-x)(-y) = xy

∴ we get , from all 4 cases

| x | ⋅ | y | = | x y |

Hence Proved.

c)

To prove - | x + y | ≤ | x | + | y |

Let

|x| = |x + y - y|

    ≥ |x + y| - |y|     ( Triangle inequality)

⇒ |x| + |y| ≥ |x + y|

Hence Proved.