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3 years ago

What is the sum of 7 and 3 sixteenths plus 3 and three fourths

Mathematics

1 answer:

aalyn [17]3 years ago

8 0

7 3/16 + 3 3/4
7 3/16+ 3 12/16
10 15/16

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C=2(3.14)r solve for r

Neporo4naja [7]

<span>C=2(3.14)r
C = 6.28r
r = C / 6.28

hope it helps</span>

7 0

3 years ago

Read 2 more answers

PLZ HELP I really need help with this.

kvv77 [185]

Answer:

$x = \frac{mp + e}{t}$

Step-by-step explanation:

times p on both sides

then plus e on both sides

then divide t on both sides

3 0

3 years ago

Expand and simplify (2x^2)(4x+5)-8x(x^2-2)

Olegator [25]

The answer would be 10x^2 + 16x

good luck on whatever it is you’re doing ^^

5 0

3 years ago

Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from

Lapatulllka [165]

<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$ </h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

$x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8$

The area of a triangle is= $\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]$

= $|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|$

= $|-54|$ = 54 square units

The length of AC = $\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}$

= $\sqrt{(2-12)^{2} +(1-8)^2}$

= $\sqrt{149}$ units

Let the perpendicular distance from B to AC be = x

According To Problem

$\frac{1}{2} \times x \times \sqrt{149} = 54$

⇔ $x =\frac{108}{\sqrt{149} }$ units

Therefore the perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$

5 0

3 years ago

Translating verbal sentences: 16 decreased by a number is 27

Julli [10]

16 - x = 27
x = -11

16 - x = 27

16 = 27 + x

16 - 27 = x

-11 = x

4 0

3 years ago