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2 years ago

Solve a real world problem that can be represented by the expression (-3)(5)+10

Mathematics

1 answer:

vovikov84 [41]2 years ago

6 0

Using PEMDAS you start with parenthesis

(-3)(5) = -3 x 5 = -15

-15 + 10 = -5

The answer is -5

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If a round plate has a radius of 6 inches, what is the area of the plate

weeeeeb [17]

113.04 about-ish I used 3.14 for pi

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2 years ago

Given that y = 8 cm and θ = 34°, work out x rounded to 1 DP.

morpeh [17]

Answer:

Step-by-step explanation:

if y is hypotonuse and x is opposite side:

sin 34=x/8

8*sin 34=x

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2 years ago

Find the midpoint of the line segment joining the points. (6, -5), (6, 3)

adoni [48]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Pls help for this question! no wrong answers pls

ankoles [38]

Answer:

41.29 cm²

Step-by-step explanation:

From the question,

Area of the shaded portion = Area of the circle - area of the square.

A' = πr²-L²...… Equation 1

Where A' = Area of the shade portion, r = radius of the circle, L = length of the square, π = pie

Given: r = 4 cm, L = 3 cm

Constant: π = 22/7.

Substitute these values into equation 1

A' = [(22/7)×4²]-3²

A' = 50.29-9

A' = 41.29 cm²

Hence the area of the shaded portion is 41.29 cm²

8 0

2 years ago

What is the most precise name for quadrilateral ABCD with vertices A(–5,2), B(–3, 5),C(4, 5),and D(2, 2)?

bonufazy [111]

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram , but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= $(\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})$

Mid point of BD= $(\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})$

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

8 0

2 years ago