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2 years ago

Solve for z -7(z-6)=-70

Mathematics

2 answers:

Dmitriy789 [7]2 years ago

8 0

Z = 16

Explanation:
-7 (z-6) = -70
-7z + 42 = -70
Z = -70 - 42 / -7
Z = 16

Stels [109]2 years ago

7 0

Answer:yes

Step-by-step explanation:

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Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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2 years ago

Yamri went to the store to buy gardening supplies. A bag of dirt was $3.99 and tulips cost 75 cents per bulb. He bought one bag

seraphim [82]

3.99 + 0.75b = 12.24
0.75b = 12.24 - 3.99
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2 years ago

Read 2 more answers

N is factor of 36,<br> n is factor of 48,<br> n>6

Bad White [126]

Answer:

n= 12

Step-by-step explanation:

12 times 2 gives you 36

12 times 4 gives you 48.

12 is both factors of these numbers

7 0

3 years ago

"Complete the square" to convert the equation of each circle to graphing form. Identify the center and the radius.

Reil [10]

Answer:

The center is (-3,2) and the radius is r=2

Step-by-step explanation:

The general equation of the given circle is

$x^2+6x+y^2-4y=-9$

Add the square of half the coefficient of the linear terms to both sides of the equation to obtain;

$x^2+6x+3^2+y^2-4y+(-2)^2=-9+3^2+(-2)^2$

$x^2+6x+9+y^2-4y+4=-9+9+4$

$x^2+6x+9+y^2-4y+4=4$

The quadratic trinomials in x and y on the left side of the equations are perfect squares.

We factor to obtain;

$(x+3)^2+(y-2)^2=4$

$(x--3)^2+(y-2)^2=2^2$

Comparing to:

$(x-h)^2+(y-k)^2=r^2$

The center is (-3,2) and the radius is r=2

8 0

2 years ago

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

6 0

2 years ago