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3 years ago

Please help !! What is the area of the trapezoid shown below? 4 5

Mathematics

2 answers:

Rashid [163]3 years ago

7 0

Answer:

the area of the trapezoid is (8 + 4).3/2 = 18

dangina [55]3 years ago

3 0

Answer:

Step-by-step explanation:

khan

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yarga [219]

$The\ slope-intercept\ form:y=mx+b\\\\m\ is\ the slope\\b\ is\ y-intercept\\\\7x+y=8\\\\y=-7x+8\\\\the\ slope:m=-7\\y-intercept:b=8$

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3 years ago

Which vectors represent the reflection of the vector <3, -7> across the x-axis? A. [3/7} B. [-3/7] C. <-3, 7> D. &lt

Alexandra [31]

Answer:

D) < 3, 7)>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given that the vector < 3 , -7 >

Given the vector reflection across the x-axis

(x,y) → (x , -y)

The vector < 3,-7> →< 3, -(-7)>

< 3,-7> →< 3, 7)>

3 0

3 years ago

Read 2 more answers

Which figure has the same area as the parallelogram shown below? A. A triangle with a base of 4 mm and a height of 20 mm B. A tr

Arisa [49]

<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of $80mm^{2}$ , according to the following formula, which works for all rectangles and parallelograms:

$A_{parallelogram}=(b)(h)$ (1)

Where $b$ is the base and $h$ is the height

The<u> area of a triangle</u> is given by the following formula:

$A_{triangle}=\frac{1}{2}(b)(h)$ (2)

So, for option A:

$A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}$

Now, the <u>area of a trapezoid </u>is:

$A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)$ (3)

For option B:

$A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}$

For option C:

$A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}$ >>>>This is the correct option!

For option D:

$A_{rectangle}=(30mm)(8mm)=240mm^{2} \neq 80mm^{2}$

<h2>Therefore the correct option is C</h2>

5 0

3 years ago

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find

Anastasy [175]

Required area of shaded portion

= Area of square ABCD - 4 × area of one quadrant

$= 14 \times 14 - 4 \times \frac{90}{360} \times \frac{22}{7} \times (7) \times (7)$

$= 196 - 154 = 42 \: cm {}^{2}$

3 0

3 years ago

what is the length of the hypotenuse of a right triangle with legs that measure 8 and 13 cm round the answer to the nearest hund

atroni [7]

Use pythagoreaan theorem
a^2+b^2=c^2
a and b are teh legnth s of the legs
c is the legnt o fthe hyptonuse

8^2+13^2=c^2
64+169=c^2
233=c^2
square root both sdies
aprox 15.264=c

the legnth is aprox 15.26 cm

4 0

3 years ago

Read 2 more answers