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3 years ago

11

Emily looks at a picture. A car in this picture is 30cm long. the scale of the drawing is 1:14

1 answer:

fenix001 [56]3 years ago

4 0

Answer:

Step-by-step explanation:

30 cm:

1: 14

so 1*14 = 14

so 30*14=420

so if you are trying to figure out how long it is in real life it's supposed to be 420 something.

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How can the x-intercept and y-intercept be used to graph a linear equation

JulijaS [17]

Because a line always goes past the y-intercept and x-intercept. It's not always both, it can sometimes just be the x-intercept or the y-intercept.

When a line intersects these points, for example if a line was to intersect the x-axis then y would be equal to 0, and the opposite for the y-axis. If a line was to intersect the y-axis x would be equal to 0.

Therefore by using that knowledge, and the equation of the line [ y=mx+c or y-y1=m(x-x1) ], we can find the equation of our line. Of course you would need the gradient of that line (the value "m").

5 0

4 years ago

0.336 divided by 0.2

Lilit [14]

1.68 I hope it helps!

7 0

3 years ago

Elza [17]

Answer:

$(b)\ 2x - 9 = -1$

$(c)\ -2x + 12 = 4$

$(e)\ 2(x - 10) = -12$

Step-by-step explanation:

Required

Which equals $x = 4$

$(a)\ 2x + 6 = 2$

Collect like terms

$2x = 2 - 6$

$2x = -4$

Divide both sides by 2

$x = -2$

$(b)\ 2x - 9 = -1$

Collect like terms

$2x = -1 + 9$

$2x = 8$

Divide both sides by 2

$x = 4$

$(c)\ -2x + 12 = 4$

Collect like terms

$-2x =- 12 + 4$

$-2x =-8$

Divide both sides by -2

$x = 4$

$(d)\ 2(x + 2) = 10$

Divide both sides by 2

$x + 2 =5$

Collect like terms

$x = 5-2$

$x = 3$

$(e)\ 2(x - 10) = -12$

Divide both sides by 2

$x - 10 = -6$

Collect like terms

$x = 10 - 6$

$x = 4$

Hence, the equations with the required solution are:

$(b)\ 2x - 9 = -1$

$(c)\ -2x + 12 = 4$

$(e)\ 2(x - 10) = -12$

3 0

3 years ago

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

3 years ago

The first four terms of a sequence are shown below:

Marta_Voda [28]

Answer:

Step-by-step explanation:

Since each term is decreased by three $f(n+1)=f(n)-3$

7 0

3 years ago