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3 years ago

Find The Sum. 6 2/3 + 1 5/6=

Mathematics

2 answers:

patriot [66]3 years ago

5 0

6 2/3 + 1 5/6 = 8 1/2

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Svetlanka [38]3 years ago

4 0

You first turn both fractions into improper fractions. then you get 20/3 and 11/6. afterthat multiply 20/3 by 2 because you need the same denominator as 11/6. so that will be 40/6. then add 40/6 and 11/6 = 51/6. Lastly turn 51/6 into a mixed number by dividing 51÷6= 8.5 meaning the answer would be 8 and 1/2

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stellarik [79]

Answer:
10^3
Step-by-step explanation:
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2 years ago

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Solve for x by completing the square: x2 - 12x + 11 = 0

sineoko [7]

Answer:

Step-by-step explanation:

x² - 12x + 11 = 0 ( subtract 11 from both sides )

x² - 12x = - 11

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 6)x + 36 = - 11 + 36

(x - 6)² = 25 ( take square root of both sides )

x - 6 = ± $\sqrt{25}$ = ± 5 ( add 6 to both sides )

x = 6 ± 5

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2 years ago

The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

Marrrta [24]

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

$GH=\sqrt{13}\ units$

step 2

Find the distance IH

substitute in the formula

$d=\sqrt{(0+1)^{2}+(9-5)^{2}}$

$d=\sqrt{(1)^{2}+(4)^{2}}$

$IH=\sqrt{17}\ units$

step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

$a=\sqrt{17}\ units$

$b=\sqrt{13}\ units$

substitute

$c^{2}= (\sqrt{20})^{2}=20$

$a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30$

therefore

$c^{2}< a^{2}+b^{2}$

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3 years ago

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kolezko [41]

Answer:

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Step-by-step explanation:

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3 years ago

17) The coordinates of quadrilateral EFGH are E(7, 7), F(4, 8), G(2, 2), H(7, 2). Show that EF is perpendicular to FG.

strojnjashka [21]

Answer:

see explanation

Step-by-step explanation:

first we calculate the slopes of EF and FG using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ = E (7, 7 ) and F (4, 8 )

$m_{EF}$ = $\frac{8-7}{4-7}$ = $\frac{1}{-3}$ = - $\frac{1}{3}$

repeat with (x₁, y₁ ) = F (4, 8 ) and (x₂, y₂ ) = G (2, 2 )

$m_{FG}$ = $\frac{2-8}{2-4}$ = $\frac{-6}{-2}$ = 3

if 2 lines are perpendicular then the product of their slopes = - 1

here

$m_{EF}$ × $m_{FG}$ = - $\frac{1}{3}$ × 3 = - 1

then EF is perpendicular to FG

3 0

2 years ago