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4 years ago

Write 6 1/3 as an improper fraction

Mathematics

2 answers:

Elenna [48]4 years ago

8 0

Answer:

19/3

Step-by-step explanation:

6 1/3 > 6*3 = 18 + 1 = 19 > 19/3

Multiply the regular number (6) by the denominator (3) then add the numerator (1) and then put that over the original denominator. :)

laiz [17]4 years ago

7 0

Answer:

Step-by-step explanation:

19/3

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Plzzz HELP (3 QUESTIONS!!)

k0ka [10]

Answer:

6: d, 7: b and c, 8: a

Step-by-step explanation:

6) The answer is (d).

We have,

0.00001131 =1.131×0.00001

= 1.131/100000

= 1.131×10^-5.

So, this is correct answer.

7) This question have two answers (b) and (c), both are correct.

(b) 480000-97000

= 4.8x100000-9.7x10000

=4.8x10^5-9.7x10^4

= 4.8x10^5-0.97x10^5

= (4.8-0.97) x10^5

=3.83x10^5

8)(a)

(4 x 10^3)(5x10^7).

= 4x5x10³x10^7

= 4x5x10^(3+7).

8 0

3 years ago

I need help to find the measures and equation pls.

siniylev [52]

Answer:

Equation: 7x + 8x = 180

x = 12

∠CBA = 84

∠CFH = 96

Step-by-step explanation:

We can see that ∠CBA = ∠CFE and ∠CBD = ∠CFH.

We know that the sums of two angles on a straight line are going to be equal to 180.

∠CBA = 7x

∠CFH = 8x

To find the value of x, we must do the following:

7x + 8x = 180

15x = 180

15x/15 = 180/15

x = 12

Now we just substitute to find the angle measures:

∠CBA = 7 · 12 = 84

∠ CFH = 8 · 12 = 96

7 0

2 years ago

Find the quotient -44a^3b^4 - 64a^4b3 - 84ab<br> ___________________<br> -4ab

Aliun [14]

When you divide a variable with an exponent by a variable with an exponent, you subtract the exponents (only if the bases/variables are the same)

For example:

$\frac{x^2}{y^2}$ Since they have different variables (x and y), you can't simplify this anymore

$\frac{x^5}{x^3} = x^{5-3}=x^2$

$\frac{z^3x^7}{z^2yx^2} = (\frac{1}{y})(z^{3-2})(x^{7-2}) = (\frac{1}{y} )(z^1)(x^5) = \frac{zx^5}{y}$

( if the exponent happens to come out negative, you have to move the variable and the exponent to the other side of the fraction to make the exponent positive)

Ex:

$\frac{x^2}{x^4} =x^{2-4}=x^{-2} = \frac{1}{x^2}$

$\frac{-44a^3b^4-64a^4b^3-84ab}{-4ab}$ SInce they all have a common factor of 4, you can divide out -4

$\frac{11a^3b^4+16a^4b^3+21ab}{ab}$

If you follow the steps from ^^^^^^^, you should get

$11a^2b^3+16a^3b^2+21$

7 0

3 years ago

Which of the triangles are right triangles?

zhannawk [14.2K]

Answer:

4. is the right triangle

Step-by-step explanation:

both sides(with numbers) says 3 IM SORRY IM BAD AT EXPLAINING but do hope this helps!

8 0

3 years ago

Read 2 more answers

Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

AlladinOne [14]

Answer:

The distance between the two parallel lines is 11/17 units

Step-by-step explanation:

we have

8x-15y+5=0 -----> equation A

16x-30y-12=0 ----> equation B

Divide by 2 both sides equation B

8x-15y-6=0 ----> equation C

Compare equation A and equation C

Line A and Line C are parallel lines with different y-intercept

step 1

Find the slope of the parallel lines (The slope of two parallel lines is the same)

8x-15y+5=0

15y=8x+5

$y=\frac{8}{15}x+\frac{1}{3}$

the slope is

$m=\frac{8}{15}$

step 2

Find the slope of a line perpendicular to the given lines

Remember that

If two lines are perpendicular then their slopes are opposite reciprocal (the product of their slopes is -1)

$m1*m2=-1$

we have

$m1=\frac{8}{15}$

therefore

$m2=-\frac{15}{8}$

step 3

Find the equation of the line perpendicular to the given lines

assume any point that lie on line A

$y=\frac{8}{15}x+\frac{1}{3}$

For $x=0$

$y=\frac{1}{3}$

To find the equation of the line we have

$point\ (0,1/3)$ ---> is the y-intercept

$m=-\frac{15}{8}$

The equation in slope intercept form is

$y=-\frac{15}{8}x+\frac{1}{3}$ -----> equation D

step 4

Find the intersection point of the perpendicular line with the Line C

we have the system of equations

$y=-\frac{15}{8}x+\frac{1}{3}$ ----> equation D

$8x-15y-6=0$ ----> $y=\frac{8}{15}x-\frac{2}{5}$ ----> equation E

equate equation D and equation E and solve for x

$\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}$

$\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}$

Multiply by 120 both sides to remove fractions

$64x+225x=40+48$

$289x=88$

$x=88/289$

Find the value of y

$y=-\frac{15}{8}(88/289)+\frac{1}{3}$

$y=-\frac{206}{867}$

the intersection point is $(\frac{88}{289},-\frac{206}{867})$

step 5

Find the distance between the points $(0,\frac{1}{3})$ and $(\frac{88}{289},-\frac{206}{867})$

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

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

substitute the values

$d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}$

$d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}$

$d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}$

$d=\sqrt{\frac{314,721}{751,689}}$

$d=\frac{561}{867}\ units$

Simplify

$d=\frac{11}{17}\ units$

therefore

The distance between the two parallel lines is 11/17 units

see the attached figure to better understand the problem

4 0

3 years ago