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Alenkasestr [34]
4 years ago
12

Plz help also show your work.✌

Mathematics
1 answer:
I am Lyosha [343]4 years ago
5 0
The answer is A because c is the amount of cookies and she ate three. You divide it by two because she only gave HALF of it away.
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Hello help me with this question thanks in advance​
Ede4ka [16]

\bold{\huge{\green{\underline{ Solutions }}}}

<h3><u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{HM = 5 cm }

  • <u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>

<u>The </u><u>perimeter </u><u>of </u><u>square </u>

\sf{ = 4 × side }

\sf{ = 4 × 5 }

\sf{ = 20 cm }

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{MX  = 3.5 cm }

  • <u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>

<u>Here</u><u>, </u>

\sf{MX  = MT/2}

\sf{MT = 2 * 3.5 }

\sf{MT = 7 cm}

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>

  • <u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>From </u><u>above </u>

\sf{\angle{MAT  = 90° }}

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>know </u><u>that</u><u>, </u>

  • <u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>Therefore</u><u>, </u>

\sf{\angle{MHA  = }}{\sf{\angle{ MHT/2}}}

\sf{\angle{MHA = 90°/2}}

\sf{\angle {MHA = 45°}}

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Both a and b

  • <u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
  • <u>The </u><u>diagonals </u><u>are </u><u>congruent </u>

Hence, Option C is correct

<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>

In rhombus PALM,

  • <u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>

Let O be the midpoint of Rhombus PALM

<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>

\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}

\sf{\angle{OLM = 180° - 125°}}

\sf{\angle{ OLM = 55° }}

<u>Now</u><u>, </u>

\sf{\angle{OLM = }}{\sf{\angle{OLA}}}

  • <u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>

<u>Therefore</u><u>, </u>

\sf{\angle{ PLA = 55° }}

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0
3 years ago
Put this equation in slope intercept form 5x-2y=10
Katen [24]
This is the answer and the steps for doing it

4 0
3 years ago
When bivariate data from a survey is graphed what is graph called ?
Vinvika [58]

Answer:

D

Step-by-step explanation:

Scatter plot (Use excel with any bivariate data you can find)

7 0
3 years ago
The quadratic function h(t) = -16.1? + 150 models a ball's height, in feet, over time, in seconds,
Sliva [168]
<h2>Hello!</h2>

The answer is: The first graphic representation.

<h2>Why?</h2>

We are given a quadratic equation, meaning that it could be two possible solutions for the exercise, however, we are talking about time, so we have to consider only the obtained positive values.

Let's make the equation equal to 0 in order to find the values of "t"

h(t) = -16.1t^2 + 150

0=-16.1t^2+150

16.1t^2=150

t^2=\frac{150}{16.1} =9.32

t=+-\sqrt{9.32} =+-3.05

t1=3.05

t2=-305

So, discarding the negative value, we can use the possitive value to find the correct graphic representation.

To find the correct graphic representation we must take into consideration the following:

- We must remember that the sign of the coefficient of the quadratic term (t^2) will define if the parabola opens downward or upward.

From the given quadratic (or parabola) equation we have:

a=-1

b=0

c=150

So, since the coefficient of the quadratic term is negative, the parabola opens downward.

- Since we are looking for a graphic that represents the change in height over time, we need to look for a graphic that shows only positive values for the x-axis (time)

- We are looking for a parabola which y-axis intercept is equal to 150.

Therefore, the graphic representation of the quadratic function that models a ball's height over time is the first graphic representation.

Have a nice day!

5 0
2 years ago
??????? Help please
Ad libitum [116K]

Step-by-step explanation:

Put the values of x = 6, y = -7 and z = 0.8 to the expressions:

a) 5x → 5(6) = 30

b) 3y → 3(-7) = -21

c) 10z → 10(0.8) = 8

4 0
3 years ago
Read 2 more answers
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