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Sati [7]
3 years ago
6

Mary Jane made 140 cupcakes for the sixth grade classes at Oak School. She frosted 3/5 of the cupcakes with caramel icing and pu

t
sprinkles on 1/2 of those cupcakes. How many caramel cupcakes had sprinkles?

A)
38 cupcakes

B)
42 cupcakes

C)
44 cupcakes

D)
46 cupcakes
Mathematics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

B). 42 cupcakes

Step-by-step explanation:

3/6= 0.6

1/2=  0.5

140*0.6=84

84*0.5=42

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What is the following in simplest form? sqrt(8)+3sqrt(2)+sqrt(32)
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\qquad 2^4\cdot 2\\
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\\\\\\
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5 0
3 years ago
Help me please no links :)
AlekseyPX

Answer:

Domain (Basically means the sets of values of x-axis).Hence, the set of values of domain is given by -3<domain<1. Meaning domain lies between values more than -3 and less than 1.

Range (Basically means the sets of values of y-axis). Hence, the set of values of range is given by -1<range<3. Meaning that range lies between values more than -1 and less than 3.

Function basically means the equation of the given circle. We know that equation of circle is given by;

(x-h)^2+ (y-k)^2 = r^2 [where (h, k) is the center of circle and r is the radius of circle]

From the figure,

(h, k)= (-1 ,1) and r= 2.

So, we can conclude that function is given by equation:

(x-(-1))^2 + (y-1)^2 =  2^2

(x+1)^2  +  (y-1)^2 = 4.

4 0
3 years ago
Solve for x. 89(54x−36)+2=−34(−40+16x)+90x
krok68 [10]
Eliminate parentheses using the distributive property.
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3 years ago
39 ft./m is how many yards per second
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4 0
3 years ago
: If tan A=- 9/13 and 0 &lt; A &lt; 180, find, without using tables or calculators, the value of
stich3 [128]

Answer:  \bold{\sin A=\dfrac{9\sqrt{10}}{50}\qquad \cos A=-\dfrac{13\sqrt{10}}{50}}

<u>Step-by-step explanation:</u>

0° < A < 180° so it is in Quadrant I or II

since tan A is negative, it is in Quadrant II ---> (-x, +y)

\text{Given}:\tan A=\dfrac{y}{x}\ =\dfrac{9}{-13}

Next, find the hypotenuse given that x = -13 and y = 9

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\sin A = \dfrac{y}{r}\quad =\dfrac{9}{5\sqrt{10}}\bigg(\dfrac{\sqrt{10}}{\sqrt{10}}\bigg)\quad =\large\boxed{\dfrac{9\sqrt{10}}{50}}\\\\\\\\\cos A = \dfrac{x}{r}\quad =\dfrac{-13}{5\sqrt{10}}\bigg(\dfrac{\sqrt{10}}{\sqrt{10}}\bigg)\quad =\large\boxed{-\dfrac{13\sqrt{10}}{50}}

8 0
3 years ago
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