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Mashutka [201]
3 years ago
14

at party the pumpkin pie is cut into 14 slices and the cherry pies cut into seven slices if the host wants to serve identical pl

ates that contained the same combination of pumpkin and Cherry sizes with no slices leftover what is the greatest number of plates the host can prepare
Mathematics
1 answer:
Lyrx [107]3 years ago
7 0

Answer:

The host can only prepare 7 plates

Step-by-step explanation:

7 plates, each will contain 1 slice of cherry pie and 2 slices of pumpkin pie. (14/7 = 2)

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Which answer is equal to the quotient in the expression below?
Gennadij [26K]
FIRST section: x^2-3x+2 = x^2-x-2x+2=(x^2-x) -2x+2 = (x^2-x) -2(x-a million) = x(x-a million) -2(x-a million) = (x-2)(x-a million) 2nd section: x^-4 = x^2- 2^2 = (x-2)(x+2) So now your equation looks like this: FIRST section / 2nd section or (x-2)(x-a million) / (x-2)(x+2) and this comes out at (x-a million) / (x+2), so the respond is B.
7 0
3 years ago
Rhombus JKLM is shown below. Find m∠KLM.
lianna [129]

Answer:

Option D

Step-by-step explanation:

Properties of a rhombus,

1). Diagonals of a rhombus bisect the opposite angles.

2). Sum of adjacent two angles of a rhombus is 180°.

By property (1),

Diagonal JL will bisect the angles ∠MJK and ∠KLM.

Therefore, m∠KLM = 2(5x + 3)°

Similarly, diagonal KM will bisect the angles ∠JKL and JML.

Therefore, m∠JKL = 2(9x - 11)°

By property (2),

2(9x - 11)° + 2(5x + 3)° = 180°

9x - 11 + 5x + 3 = 90

14x - 8 = 90

14x = 98

x = 7

Since, m∠KLM = 2(5x + 3)°

                        = 2(5×7 + 3)°

                        = 76°

Therefore, Option D will be the answer.

7 0
2 years ago
The weight of people in a small town in missouri is known to be normally distributed with a mean of 154 pounds and a standard de
ad-work [718]
If a random sample of 20 persons weighed 3,460, the sample mean x-bar would be 3460/20 = 173 pounds.
The z-score for 173 pounds is given by:
z=\frac{173-154}{29}=0.655
Referring to a standard normal distribution table, and using z = 0.66, we find:
P(\bar x\ \textless \ 173)=0.7454
Therefore
P(\bar x\ \textgreater \ 173)=1-0.7454=0.2546
The answer is: 0.2546

4 0
3 years ago
What is the area of the figure below?
murzikaleks [220]
96m^2 is your answer

6 0
3 years ago
Read 2 more answers
How to solve the second equation
ivann1987 [24]
\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
y = &{{ -0.02}}x^2&{{ +1}}x&{{ +6}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{1^2}{2(-0.02)}~~,~~6-\cfrac{1^2}{4(-0.02)} \right)
\\\\\\
(\stackrel{\textit{how far ahead it went}}{25}~,~\stackrel{\textit{how high it went}}{18\frac{1}{2}})

\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
y = &{{ -0.01}}x^2&{{ +0.7}}x&{{ +6}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{0.7^2}{2(-0.01)}~~,~~6-\cfrac{0.7^2}{4(-0.01)} \right)
\\\\\\
(\stackrel{\textit{how far ahead it went}}{24\frac{1}{2}}~,~\stackrel{\textit{how high it went}}{18\frac{1}{2}})
4 0
3 years ago
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