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3 years ago

1/8, 2/7, 1/2, 4/5, , ____

Mathematics

1 answer:

Masteriza [31]3 years ago

3 0

Answer:

5/4 and 6/3

Step-by-step explanation:

Your taking one away from denominator and adding it to numerator

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If f (x) = -9x - 9 and g (x) = Vx - 9, what is (f ° g) (10)?

Temka [501]

Answer: $(f \circ g) (10)= -18\ .$

Step-by-step explanation:

Given: $f (x) = -9x - 9$ and $g (x) = \sqrt{x - 9}$

To find : (f o g) (10)

For this we first find (f o g) (x)= $f(g(x))$

$=f(\sqrt{x-9})\\\\=-9(\sqrt{x-9})-9$

Now,

$(f \circ g) (10)=-9(\sqrt{10-9})-9\\\\=-9\sqrt{1}-9\\\\=-9-9=-18$

Hence, the value of $(f \circ g) (10)= -18\ .$

8 0

3 years ago

Expand and simplify (x-4)(x-1)

andre [41]

Step-by-step explanation:

(x - 4)(x - 1)

= x² - x - 4x + 4

= x² - 5x + 4.

6 0

3 years ago

Read 2 more answers

(picture) Factoring Polynomials: GCF PLEASEE HELP!!!!!<br> a<br> b<br> c<br> d

kramer

For this case we must find the greatest common factor of both expressions, the GFC will be the greatest number that is a factor of both numbers, that is:

The number 4 is the greatest number that divides 32 and 12 resulting in a whole number.

$\frac {32} {4} = 8\\\frac {12} {4} = 3$

In addition, both expressions have as a common factor, $a ^ 2$ :

$4a ^ 2 (8a + 3)$

Thus, $32a ^ 3 + 12a ^ 2 = 4a ^ 2 (8a + 3)$

Answer:

$4a ^ 2 (8a + 3)$

Option A

3 0

3 years ago

Find an inequality that has this graph.

Sidana [21]

Answer:

me sacas de onda

Step-by-step explanation:

eres tan bueno que nunca te equivocaras

7 0

3 years ago

Find the volume of the given solid. Bounded by the cylinder y2 + z2 = 16 and the planes x = 2y, x = 0, z = 0 in the first octant

erastovalidia [21]

Answer:

$\mathbf{ \dfrac{128}{3}}$

Step-by-step explanation:

Given that:

$y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0$

$Replace \ z = 0 \ \text{in the given equation ;} y^2+z^2=16$

$Then;$

$y^2 = 16 \\ \\ y = \sqrt{16} \\ \\ y = \pm 4$

$\text{So; in 1st octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y$

∴

$\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy \\ \\ = \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\ V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy$

$Now, let:\\\\ 16 - y^2 = u \\ \\ -2ydy = du \\ \\ 2ydy = -du \\ \\ Hence, when \ y = 0; u = 16 \\ \\ and \\ \\ when \ y = 4 \ then \ u = 0$

∴

$V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy \\ \\ = \int ^0_{16} - \sqrt{u}\ du \\ \\ = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du \\ \\ = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg] \\ \\$

$= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )$

$= - \dfrac{2}{3} \Big(64\Big )$

∴

$\mathbf{V = \dfrac{128}{3}}$

5 0

3 years ago

1/8, 2/7, 1/2, 4/5, ________, ____________​

1/8, 2/7, 1/2, 4/5, , ____