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3 years ago

How many solutions does the equation 6z - 3z - 7 = -2 + 3 have?

Mathematics

1 answer:

erik [133]3 years ago

6 0

Answer:

One

Step-by-step explanation:

6z - 3z - 7 = -2 + 3

Combine like terms

3z-7 = 1

Add 7 to each side

3z-7+7 = 1+7

3z=8

Divide each side by 3

3z/3 = 8/3

z = 8/3

There is one solution

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Simplify this expression: 19 – (–8) – (–14) = ? A. –7 B. –3 C. 41 D. 25

Mrac [35]

The answer is C. 41.

6 0

4 years ago

What is the area? Please explain steps so I can attempt to understand

yarga [219]

Answer:

$12 \sqrt{2}$

Step-by-step explanation:

you can use hypotenuse formula

45-45-90 isosceles right triangle

so, 45= x

x² + x² = 24²

2x² = 24²

x²= 24.12

x= $\sqrt{288} = 12 \sqrt{2}$

or :

you can use this rule: if you see a 45 45 90 isosceles right triagngle (you can look picture) 45=a cm and 90 = $a\sqrt{2}$

so ;

$a \sqrt{2} = 24$

a= $12 \sqrt{2}$

hope this helps ^-^

3 0

4 years ago

Factor −6 out of −18a−30

fgiga [73]

Remark

This is like using the distributive principle only you are going backwards. Instead of trying to remove brackets, you are trying to create them.

Solution

Since you want to take - 6 outside the brackets you should divide both terms by - 6

- 18a: - 18a/- 6 = 3a which is left inside the brackets.

-30: - 30/-6 = 5

- 6(3a + 5)

Answer

-6(3a + 5)

6 0

4 years ago

Read 2 more answers

Area of a circle R = 5cm Area= REMEMBER TO ANSWER TO 2 DECIMAL PLACES.

Musya8 [376]

Answer:

78.52 cm sq.

Step-by-step explanation:

Area = pi x r^2

A = pi x (5)^2

A = pi x 25

A = 3.141 x 25

A = 78.52 cm sq.

4 0

4 years ago

WILL GIVE BRAINLIEST!

Nataly [62]

Answer:

The equation the circle is given by

${\bf 4x}^{\bf 2}+{\bf 4y}^{\bf 2}-{\bf 8x+0y-60}={\bf 0}$

Step-by-step explanation:

The general form of the circle is

$x^2+y^2+Dx+Ey+F=0\hfill (1)$

Given points are (5,0) , (0,4) and (2,4)

Let the points (5,0) , (0,4) and (2,4) be (x_1,y_1) , (x_2,y_2) and (x_3,y_3) respectively

The equation of the circle passes through given points (5,0) , (0,4) and (2,4)

Substituting the point (5,0) in (1)

$(-5)^2+(0)^2+D(5)+E(0)+F=0$

$25-5D+F=0$

$5D-F=25\hfill (2)$

Substituting the point (0,4) in

$(0)^2+(4)^2+D(0)+E(4)+F=0$

$16+4E+F=0$

$4E+F=-16\hfill (3)$

Substituting the point (2,4) in

$(2)^2+(4)^2+D(2)+E(4)+F=0$

$4+16+2D+4E+F=0$

$2D+4E+F=-20\hfill (4)$

By solving the equations (2) ,(3) and(4) we get the values of D,E and F as below

$D=-2$ , $E=0$ and $F=-15$

substituting the values of D,E and F in (1) we get

$x^2+y^2-2x+(0)y-15=0$

Now multiplying the above equation into 4 we get

$4x^2+4y^2-8x+(0)y-60=0$

6 0

3 years ago