Ask question

Ask question

All categories

mihalych1998 [28]

3 years ago

11

Please Someone Help me!!

1 answer:

EastWind [94]3 years ago

3 0

(27 1/4) / (2 1/2) =
(109/4) / (5/2) =
109/4 * 2/5 =
218/20 =
10.9....so there is 10 full 2 1/2 oz bags

each bag sells for 4.25.....so 10 bags would sell for 4.25(10) = 42.50

You might be interested in

PLEASE HELP ME FOR BRAINLIEST!!!!

lana [24]

Use cosine rule and sine rule

3 0

3 years ago

Graph the image of this figure after a dilation with a scale factor of 1/3 centered at the origin.

posledela

Answer:

see attached diagram

Step-by-step explanation:

A dilation with a scale factor of $\dfrac{1}{3}$ centered at the origin has a rule:

$(x,y)\mapsto\left(\dfrac{x}{3},\dfrac{y}{3}\right).$

Then the image of the triangle ABC is triangle DEF(see attached diagram) with coordinates

$D\left(\dfrac{-3}{3},\dfrac{3}{3}\right)\Rightarrow D(-1,1);$
$E\left(\dfrac{3}{3},\dfrac{9}{3}\right)\Rightarrow E(1,3);$
$F\left(\dfrac{6}{3},\dfrac{6}{3}\right)\Rightarrow F(2,2).$

3 0

3 years ago

G(-5) = -4*|(-5)-1|

Mamont248 [21]

There is one answer, -24.

While yes, absolute value does indicate both a positive and negative will work, for every x value there will be only one y value associated with it; if there were more, then it would not meet the criteria to be a function.

7 0

3 years ago

What is 5 2/3 as a improper fraction

Sati [7]

17/3. 5 times 3 is 15. Add the 2 from the numerator to get 17/3

5 0

3 years ago

Read 2 more answers

Circle O, with center (x, y) passes through the points A(0, 0), B(-3, 0), and C(1, 2). Find the coordinates of the center of the

KonstantinChe [14]

Answer:

The center of the circle is

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

Step-by-step explanation:

Let the equation of the circle be

$x^2+y^2+2ax+2by+c=0$ , where (-a,-b) is the center of this circle.

The points lying the circle must satisfy the equation of this circle.

A(0,0)

We substitute this point to get;

$0^2+0^2+2a(0)+2b(0)+c=0$

$\implies c=0$

B(-3,0)

$(-3)^2+0^2+2a(-3)+2b(0)+c=0$

$\implies 9+0-6a+0+c=0$

$\implies -6a+c=-9$

But c=0

$\implies -6a=-9$

$\implies a=\frac{3}{2}$

C(1,2)

$1^2+2^2+2a(1)+2b(2)+c=0$

$1+4+2a+4b+c=0$

$2a+4b+c=-5$

Put the value of 'a' and 'c' to find 'b'

$2(\frac{3}{2})+4b+0=-5$

$3+4b+0=-5$

$4b=-5-3$

$4b=-8$

b=-2

Hence the center of the circle is

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

8 0

3 years ago