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3 years ago

What is the three equivalent fractions for 6/1

Mathematics

1 answer:

Kaylis [27]3 years ago

8 0

12/2, 18/3, 36/6 that is 3 equal fractions. just multiply both parts of the fraction by the same #

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Which is a counterexample that shows the statement is false? If you square a number, then the answer is greater than the number.

Iteru [2.4K]

Answer:

A. ( $\frac{1}{2}$ )² = $\frac{1}{2}$ × $\frac{1}{2}$ = $\frac{1}{4}$

Step-by-step explanation:

Answer choice (A) is the counterexample because squaring number $\frac{1}{2}$ results into number $\frac{1}{4}$ which is less than $\frac{1}{2}$ .

7 0

2 years ago

4x^2=27x+40<br> Solve by factoring

worty [1.4K]

Answer: hewo, there! your answer is below

x= -5/4

or x= 8

Step-by-step explanation:

step 1: Subtract 27x+40 from both sides.

4x2−(27x+40)=27x+40−(27x+40)

4x2−27x−40=0

Step 2: Factor left side of equation.

(4x+5)(x−8)=0

Step 3: Set factors equal to 0.

4x+5=0 or x−8=0

hope this helps you

have a great Day

Plz makr branilest

8 0

2 years ago

A linear function has a slope of 8/3 and passes through the point (0,5). What is the equation of the line?

MArishka [77]

This problem already gave us the slope and y-intercept, so all that's left to do is plug those two values into slope-intercept form.

Slope-Intercept Form: y = mx + b

---m is the slope

---b is the y-intercept

y = 8/3x + 5

Hope this helps!! :)

3 0

2 years ago

Read 2 more answers

Please help hurry!!!!!!

d1i1m1o1n [39]

Answer:

136

Step-by-step explanation:

8 0

2 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago