HELP FAST TO WHOEVER CAN PLEASE !!

Given:

The two numbers are [tex]-\dfrac{1}{2}[/tex] and [tex]-\dfrac{1}{3}[/tex].

To find:

The 4 rational numbers between the given numbers.

Solution:

We have,

[tex]-\dfrac{1}{2}[/tex] and [tex]-\dfrac{1}{3}[/tex]

First we need to make common denominators. So multiply and divide the first fraction by 3 and second fraction by 2.

[tex]-\dfrac{1\times 3}{2\times 3}=-\dfrac{3}{6}[/tex]

[tex]-\dfrac{1\times 2}{3\times 2}=-\dfrac{2}{6}[/tex]

We need to find 4 rational numbers between the given numbers. So, multiply and divide both fractions by five, (4+1=5).

[tex]-\dfrac{3\times 5}{6\times 5}=-\dfrac{15}{30}[/tex]

[tex]-\dfrac{2\times 5}{6\times 5}=-\dfrac{10}{30}[/tex]

Now, the four numbers between -15 to -10 are -14, -13, -12, -11.

[tex]-15<-14<-13<-12<-11<-10[/tex]

[tex]-\dfrac{15}{30}<-\dfrac{14}{30}<-\dfrac{13}{30}<-\dfrac{12}{30}<-\dfrac{11}{30}<-\dfrac{10}{30}[/tex]

[tex]-\dfrac{1}{2}<-\dfrac{14}{30}<-\dfrac{13}{30}<-\dfrac{12}{30}<-\dfrac{11}{30}<-\dfrac{1}{3}[/tex]

Therefore, the 4 rational number between the given numbers are [tex]-\dfrac{14}{30},\ -\dfrac{13}{30},\ -\dfrac{12}{30}\ -\dfrac{11}{30}[/tex].

Answer:

x = 6 and y = 0

Step-by-step explanation:

x + 2y = 6

x = 6 - 2y                          equation(i)

substitute the value of x

x + 2y = 6

6 - 2y - 2y = 6

-4y = 6 - 6

y = 0/-4

y = 0

substitute the value of x in equation (i)

x = 6 - 2y

x = 6 - 2*0

x = 6-0

x = 6

Answer:

221cm squre

Step-by-step explanation:

Answer: 0.0033

Step-by-step explanation:

Let x be a random variable that denotes the weights of cows.

Given: [tex]\mu = 483,\ \sigma=70[/tex]

maximum weight can be hold= 2840 pounds.

Mean weight = [tex]\frac{2840}{5}[/tex] = 568 pounds

The probability their total weight will be over the maximum allowed of 2840

= [tex]P(X>2840)[/tex]

[tex]P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{568-483}{\dfrac{70}{\sqrt{5}}})\\\\=P(z>2.715)\\\\=1-P(z<2.715)\\\\=1-0.9967=0.0033[/tex]

Hence, the required probability = 0.0033

Answer:

[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]

Step-by-step explanation:

The given expression is :

[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]

We need to prove that LHS is equal to RHS.

Taking LHS,

[tex]=9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}[/tex]

[tex]=(3)^{2\times\dfrac{3}{2}}-3-(\dfrac{1}{9})^{2\times \dfrac{-1}{2}}\\\\=27-3-9\\\\=15\\\\=RHS[/tex]

Hence, LHS = RHS.

Answer:

x = 8

In all the boxes, you will first subtract 8 with 2, which will be 6. Then you will add them.

6 + 6 + 6 = 18

Answer:

A

Step-by-step explanation:

For a number to be written in scientific notation, you need to have it in the form:

a x 10^n

Where a is between 1 and 10(not including 10), and n is an integer. We can notice that in option A, 0.1(a) is not between one and ten, so it is not in scientific notation.

Answer:

$405

Step-by-step explanation:

SI=PTR/100

=9*1*750/100

=$ 67.50

In 6 yrs= 67.50*6

=$405

Answer:

The answer is $1087.50

I got it wrong because someoneeeeeee didnt give the right answer (2 people to be exact)

Answer:

3x + 6 < 21

subtract 6 on both sides

3x < 15

reduce by dividing with 3

x < 5

remember to handle inequalities just like equations

Answer:

Step-by-step explanation:

The balloon was released at 10 yards, and it landed at 16 yards. The maximum  height the balloon reached was 27 yards. To find how high the balloon is per yard we do 27 ÷ 3 = 9

so on the table we do:

x-axis      y-axis

10        |     0

11         |     9

12        |     18

13        |     27

14        |     18

15        |     9

16        |     0

Answer:

269.0

Step-by-step explanation:

Trigonmetric area formula: 0.5 * a* b* sin(C)

a = 46

b=19

c = 38 degrees

0.5 * 46 * 19 * sin(38)

Pulg it into a calculator and you get 269.0441

Then round to the nearest tenth to get 269.0

Answer:

37.6 cm

Step-by-step explanation:

5.6m = 560cm

3.8m = 380cm

560/50 = 11.2 cm

380/50 = 7.6 cm

11.2 + 11.2 + 7.6 + 7.6 = 37.6 cm

Answer:

to blury

Step-by-step explanation:

Answer:

9.9 inches

Step-by-step explanation:

The area of a triangle is given by

A = 1/2 bh  where b is the length of the base and h is the length of the height

49 = 1/2 x*x

Multiply each side by 2

2*49 = x^2

98 = x^2

Take the square root of each side

sqrt(98) = sqrt(x^2)

9.899494937 =x

Rounding to the nearest tenth

9.9 =x

Answer:

i think it will be chose 3 times

Step-by-step explanation:

im not sure thats my pridiction

Answer:

0.4

Step-by-step explanation:

Answer:

I think its Your Values, but try and a get a second opinion

Answer:

D. 6, 6[tex]\sqrt{3}[/tex], 12

Step-by-step explanation:

the Pythagorean Theorem states that [tex]a^{2} +b^{2} =c^{2}[/tex]

so,

[tex]6^{2} =36\\(6\sqrt{3}) ^{2} =108\\12^{2} =144\\36+108=144[/tex]

Thanks! But what is it for??

Answer:

Thanks I got my free thing from this code that points out where to redeem it

Step-by-step explanation:

Answer:

area: 454.2 sq.cm

perimeter: 77.7 cm

Step-by-step explanation:

to find area, find the area of all the separate shapes and add them together:

semi-circle:

radius is 12 cm

area = [tex]\frac{\pi (12)^2\\}{2}[/tex] = 226.2 sq.cm

separate the trapezoid into two right triangles and a rectangle.

area for triangles:

[tex]\frac{1}{2}[/tex](5)(12) = 30

30 x 2 = 60 sq.cm (since there are two triangles)

area for rectangle:

12 x 14 = 168 sq.cm

area of whole shape: 226.2 + 60 + 168 = 454.2 sq.cm

for perimeter, just add the length of the exterior sides together.

we have some unknown sides, such as the semicircle and the sides of the trapezoid

we can find the hypotenuse of the triangles we used earlier to find the sides of the trapezoid with pythagorean theorem

[tex]5^2 + 12^2 = c^2[/tex]

[tex]25 + 144 = c^2[/tex]

[tex]169 = c^2[/tex]

[tex]\sqrt{169} =\sqrt{c^2}[/tex]

c = 13 cm

to find the outside perimeter of the semi-circle, solve as if you were finding the circumference of a normal circle and divide it in half

[tex]\frac{2\pi12}{2}[/tex] = 37.7 cm

now add it all up:

14 + 13 + 13 +37.7 = 77.7

For all the angles, you will have to solve for x

A: 162 = 3x

162/3 = x

54 = x

M: 120 + 2x = 180

2x = 180 - 120

2x = 60

x = 60/2

x = 30

W: 150 = 30x

150/30 = x

5 = x

S: 20 + 2x = 90

2x = 90 - 20

x = 70/2

x = 35

Q : 45 + 5x = 90

5x = 90 - 45

x = 45/5

x = 9

P: not too sure if its 150,

150 + (x+5) = 180

x+5 = 180- 150

x = 30 - 5

x = 25

now you will put all the x values in order. ( i think thats what you supposed to do)

it will be: 5, 9, 25, 30, 35, 54

hope this helps♡♡♡

Answer:

on seconded I will come back and edit my answer

Answer:

3x

Step-by-step explanation:

3x^2 - 6x / x - 2

3x(x - 2) / x-2

cancels out

3x

Answer:

[tex]11.2[/tex]

Step-by-step explanation:

We can use the distance formula to solve this problem.

This distance formula states that the distance between two points [tex](x_1, y_1)[/tex] and [tex](x_2,y_2)[/tex] is equal to [tex]\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex].

School is at coordinate point (9,6) and the closest point to school on Cherry St. is (4,-4).

Thus, the distance between school and Cherry Street is [tex]\sqrt{(9-4)^2+(6-(-4))^2}=\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt{5}\approx \boxed{11.2}[/tex]

Answer:

Total cost of 6 bags of popcorn and 3 drinks: [tex]\$39.75[/tex]

Total cost of [tex]p[/tex] bags of popcorn and [tex]d[/tex] drinks: [tex]4.75p+3.75d[/tex]

Step-by-step explanation:

Information given:

Therefore, the total cost of 6 bags of popcorn and 3 drinks must be [tex]6\cdot 4.75+3\cdot 3.75=\boxed{\$39.75}[/tex]

The cost of [tex]p[/tex] bags of popcorn and [tex]d[/tex] drinks is then [tex]\boxed{4.75p+3.75d}[/tex]

Answer:

(-7,-1)

Step-by-step explanation:

bcoz coordinates of a is (-6,-2) and x=-6 we use the x-1 to find the new x which is

-6-1=-7

and to find y you should use y-3 and replace y with 2 which will be

2-3=-1

and y =-1

so the new coordinates are -7 for x-axis and -1for y-axis

Answer:

[tex]\text{C. }g(x)=(x-5)^2[/tex]

Step-by-step explanation:

The function [tex]y=(x-c)^2[/tex] has a phase shift of [tex]c[/tex] to the parent function [tex]y=x^2[/tex]. Since [tex]g(x)[/tex] has moved to the right 5 units from the parent function [tex]f(x)=x^2[/tex], we're looking for a value of [tex]c=5[/tex].

Therefore, we have:

[tex]g(x)=\boxed{(x-5)^2}[/tex]