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2 years ago

i buy 3 apples to 4 bananas. If I buy 6 more bananas than apples, how many total pieces of fruit did I buy?

Mathematics

1 answer:

ASHA 777 [7]2 years ago

8 0

Answer:

Step-by-step explanation:

The difference in ratio units is 1, which corresponds to a difference in fruit count of 6. So the total of 7 ratio units must represent 7×6 = 42 pieces of fruit.

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Please help I will make Brainliest if correct (show working)

Ann [662]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

Solve the equation or formula for the variable specified 8qr+4t=7u for q ?

g100num [7]

Answer:

q= −4t+7u over 8r

Step-by-step explanation:

8qr+4t=7u

Step 1: Add -4t to both sides.

8qr+4t+−4t=7u+−4t

8qr=−4t+7u

Step 2: Divide both sides by 8r.

8qr

−4t+7u

Answer:

−4t+7u

4 0

2 years ago

A garden is 14 1/2 ft by 13 1/4 ft. What is the area

Marianna [84]

The area means to multiply the length by the width.

let's solve.

14 1/2 x 13 1/4.
convert our fractions into decimals.

14 1/2 = 14.50
13 1/4 = 13.25

Now we can multiply properly.

14.50 x 13.25 = 192.125
since our shape is 2-Dimensional, our units are ^2.

Your area is:
192.25 ft^2

3 0

2 years ago

PLEASE HELP!! Find the coordinates.

Mazyrski [523]

Answer:

Option (D)

Step-by-step explanation:

Coordinates of the points J, E and V are,

J → (-4, -5)

E → (-4, -3)

V → (-1, -1)

This triangle is translated by the rule $T_{}$ given in the question.

Coordinates of the image will follow the rule,

(x, y) → [(x + 2), (y + 4)]

following this rule coordinates of the image triangle will be,

J(-4, 5) → J'(-2, -1)

E(-4, -3) → E'(-2, 1)

V(-1, -1) → V'(1, 3)

Therefore, points given in the option (D) will be the answer.

7 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

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

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

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

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

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

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago