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Alenkasestr [34]

3 years ago

8

Please help!! Picture included!

1 answer:

elena-s [515]3 years ago

8 0

Answer: the answer is c

Step-by-step explanation:brainlist $\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

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Jeremy used 11 gallons and 3 quarts of water to water his garden. How many quarts of water did Jeremy use?

Anon25 [30]

Answer:

47

Step-by-step explanation:

since each gallon is 4 quarts you would multiply by 11. 44 quarts then add 3 extra quarts to get 47

6 0

2 years ago

Todd is 81 years younger than Rosa. 3 years ago, Rosa’s age was 4 times Todd’s age. How old is Todd now?

Delvig [45]

Let t represent Todd's age now.
.. 4(t -3) -(t -3) = 81 . . . . . . 3 years ago, their differnce in ages was 81.
.. 3t -9 = 81
.. t = (81 +9)/3 = 30

Todd is 30 now.

_____
You can also work this by considering "ratio units." 3 years ago, the ratio of their ages was 4:1, a difference of 3. That difference corresponds to 81 years, so each "ratio unit" represents 81/3 = 27 years. Todd's age then was 1 ratio unit, 27 years. Now, Todd's age is 30.

8 0

3 years ago

Vanessa is solving the equation -3 + 4x=9

mina [271]

Answer:

The -3

Step-by-step explanation:

Hope this helps :)

8 0

3 years ago

Read 2 more answers

Write an equation of the line satisfying the given conditions. Through (2, -5), parallel to 5x 6y + 7

mojhsa [17]

We have to determine the equation of the line passing through the point (2,-5) and parallel to the line $5x = 6y+7$

When two lines are parallel, then the slopes of the two lines are equal.

Equation of line with point $(x_1, y_1)$ and slope 'm' is given by:

$(y-y_1) = m(x-x_1)$

Since, we have to determine the equation of a line with point (2,-5).

So, the equation of the line is : $(y-(-5)) = m(x-2)$

$y+5 = m(x-2)$

Since, the line is parallel to the line $5x = 6y+ 7$

So, $6y +7 = 5x$

$6y = 5x - 7$

$y = \frac{5}{6}x - \frac{7}{6}$

So, slope of the line 'm' is $\frac{5}{6}$ .

Therefore, the equation of the line is:

$(y+5) = \frac{5}{6}(x-2)$

$6y + 30 = 5x - 10$

$6y - 5x = -10-30$

$6y - 5x = -40$

Therefore, $6y - 5x = -40$ is the required equation of the line.

8 0

3 years ago

Using the method of completing the square, put each circle into the form

tatiyna

Answer:

Standard form: $(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Center: $(\frac{1}{2}, 0)$

Radius: $r =\sqrt{15}$

Step-by-step explanation:

The equation of a circle in the standard form is

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

Where the point (h, k) is the center of the circle

To transform this equation $4x^{2} -4x + 4y^{2} - 59 = 0$ this equation in the standard form we use the method of square.

First, we group similar variables

$(4x^{2} -4x) + (4y^{2}) - 59 = 0$

Divide both sides of equality by 4

$(x^{2} -x) + (y^{2}) - 14.75 = 0$

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

$b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}$

Now add $(\frac{b}{2})^2$ on both sides of the equality

$(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})$

Factor the expression and simplify the independent terms

$(x-\frac{1}{2})^2 + (y^{2}) = 15$

$(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Then

$h =\frac{1}{2}\\\\k=0$

and the center is $(\frac{1}{2}, 0)$

radius $r =\sqrt{15}$

3 0

3 years ago