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3 years ago

List 3 values that would make this inequality true. 65<15x

Mathematics

2 answers:

Eddi Din [679]3 years ago

6 0

3 values for x that would make this inequality true are anything that is greater that 4.3 like 6, 8, 11, etc.

Stolb23 [73]3 years ago

6 0

65<15X6 if that is what you'er looking for

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What two rational expressions sum to <img src="https://tex.z-dn.net/?f=%5Cfrac%7B4x%2B2%7D%7Bx%5E%7B2%7D-9%2B8%20%7D" id="TexFor

bulgar [2K]

Answer:

$\frac{4x+2}{x^2 - 9x + 8} = \frac{4x}{(x-8)(x-1)} + \frac{2}{(x-8)(x-1)}$

Step-by-step explanation:

Given

$\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}$

Required

Fill in the gaps

Going by the given parameters, we have that

$\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}$

$x^2 - 9x + 8$ , when factorized is $(x-1)(x-8)$

Hence; the expression becomes

$\frac{4x+2}{(x-1)(x-8)} = \frac{A}{(x-8)(x-1)} + \frac{B}{(x-1)(x-8)}$

Combine Fractions

$\frac{4x+2}{(x-1)(x-8)} = \frac{A + B}{(x-8)(x-1)}$

Simplify the denominators

$4x + 2 = A + B$

By direct comparison

$A = 4x$

$B = 2$

Hence, the complete expression is

$\frac{4x+2}{x^2 - 9x + 8} = \frac{4x}{(x-8)(x-1)} + \frac{2}{(x-8)(x-1)}$

3 0

3 years ago

Read 2 more answers

One basketball weighs 20.2 ounces. The basketball team has a total of 15 basketballs. If each basketball weighs the same, about

S_A_V [24]

So u will do 20.2*15=303 so ure answer would be 303 oz

6 0

2 years ago

Read 2 more answers

Solve this please thankyou

Bond [772]

Answer:

Step-by-step explanation:

-7 is < than 6 so we use the top equation

(-3 multiplied by -7) +1

21+1

22 is the answer

6 0

2 years ago

Find the tenth term of the expansion (x+ y)¹³

Anon25 [30]

Answer:

$715x^4y^9$

Step-by-step explanation:

Given

$(x + y)^{13$

Required

Determine the 10th term

Using binomial expansion, we have:

$(a + b)^n = ^nC_0a^nb^0 + ^nC_1a^{n-1}b^1 + ^nC_2a^{n-2}b^2 +.....+^nC_na^{0}b^n$

For, the 10th term. n = 9

So, we have:

$(a + b)^n = ^nC_{9}a^{n-9}b^{9$

$(x + y)^{13} = ^{13}C_{9}x^{13-9}y^9$

$(x + y)^{13} = ^{13}C_{9}x^4y^9$

Apply combination formula

$(x + y)^{13} = \frac{13!}{(13-9)!9!}x^4y^9$

$(x + y)^{13} = \frac{13!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10*9!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4*3*2*1}x^4y^9$

$(x + y)^{13} = \frac{17160}{24}x^4y^9$

$(x + y)^{13} = 715x^4y^9$

Hence, the 10th term is $715x^4y^9$

3 0

2 years ago

Solve the system and round to the nearest hundredth. Enter the

hram777 [196]

Answer:

Step-by-step explanation:

xy = 300

x + y = 50 Solve for y

y = 50 - x substitute into xy = 300

x(50 - x) = 300 Remove the brackets.

50x - x^2 = 300 Bring the left to the right.

0 = x^2 - 50x + 300

This is a quadratic. It will have 2 solutions.

a=1

b = - 50

c = 300

Put these into the quadratic equation.

It turns out that x has two values -- both plus

x1 = 42.03

x2 = 6.97

x1 + y1 = 50

42.03 + y = 50

y = 50 - 42.03

y = 7,97

(42.03 , 7.97)

====================

x2 + y2 = 50

6.97 + y2 = 50

y2 = 50 - 6.97

y2 = 43.03

(6.97 , 43.03)

6 0

2 years ago

Read 2 more answers