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4 years ago

Find four consecutive even integers whose sum is 144

1 answer:

5 0

Ok so consecutive even numbers are 2 apart, like 2,4,6

so they are
x,x+2,x+4,x+6

total is 144
x+x+2+x+4+x+6=144
4x+12=144
minus 12 both sides
4x=132
divide both sides by 4
x=33
impossible, why is it odd
if it was 148 or 140, then we could have a sum of four consecutive integers, but no, so answer is no solution unless you mistyped and put even instead of odd or something

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What is the value of (16^3/2)^1/2 256 8 4096 64

BabaBlast [244]

Answer:

Step-by-step explanation:

(16^3/2)^1/2

We know that a^ b^c = a^(b*c)

16 ^(3/2*1/2)

16 ^3/4

Now rewriting 16 as 2^4

2^4^3/4

2 ^ (4*3/4)

2^3

6 0

4 years ago

Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers

OleMash [197]

Answer:

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

Step-by-step explanation:

From Sine rule

$\frac{a}{sinA}=\frac{b}{sinB} = \frac{c}{sinC}$

∴ b / sinB = c / sinC

From the question,

b = 129, c = 168 and ∠B = 48°

∴ 129 / sin48° = 168 / sinC

Then, sinC = (168×sin48)/129

sinC = 0.9678

C = sin⁻¹(0.9678)

C = 75.42

∠C2=75.4°

and

∴∠C1 = 180° - 75.4°

∠C1 =104.6°

For ∠A

∠A1 = 180° - (104.6°+48°) [sum of angles in a triangle]

∠A1 = 27.4°

and

∠A2 = 180° - (75.4° + 48°)

∠A2 = 180° - (123.4°)

∠A2 = 56.6°

For side a

a1/sinA1 = b/sinB

a1/ sin27.4° = 129/sin48

a1 = (129×sin27.4°)/sin48

a1 = 79.8845

a1 = 79.9

and

a2/sinA2 = b / sinB

a2/ sin56.6° = 129/sin48

a2 = (129×sin56.6°)/sin48

a2 = 144.9184

a2 = 144.9

Hence,

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

4 0

3 years ago

Help me I sorta get it but still help

lara31 [8.8K]

Answer:

ok so u jus gotta pick all that apply or is equal to -1/2(4 - 2 + 6x). i hope this helpsssssssssss

Step-by-step explanation:

5 0

3 years ago

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8.<br><br> Find x and why please

lina2011 [118]

Use photomath lol it’s quick with those

6 0

3 years ago

Someone just answer this math question thingie lol. you’ll get 33 points G

Stels [109]

1) $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Step-by-step explanation:

We need to solve the polynomials.

1) $(-2x+4)+(x-6)$

Solving:

$(-2x+4)+(x-6)\\=-2x+4+x-6\\Combining\,\,like\,\,terms:\\=-2x+x+4-6\\=-x-2$

So, $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$

Solving:

$(uv-3)+(4uv+1)$

Expanding:

$=uv-3+4uv+1\\Combining\,\,like\,\,terms:\\=uv+4uv-3+1\\=5uv-2\\$

So, $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$

Solving:

$(11x-8)-(2x+3)$

Expanding:

$=11x-8-2x-3\\Combining\,\,like\,\,terms:\,\,\\=11x-2x-8-3\\=-9x-11$

So, $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$

Solving:

$(x^2-4)-(x-2)$

Expanding:

$=x^2-4-x+2\\Combining\,\,like\,\,terms:\\=x^2-x-4+2\\=x^2-x-2$

SO, $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$

Solving:

$(12-3t-7t^2)+(1+3t-t^2)$

Expanding:

$=12-3t-7t^2+1+3t-t^2\\Combining\,\,like\,\,terms:\\=12+1-3t+3t-7t^2-t^2\\=13+0t-8t^2\\=-8t^2+13$

So, $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Keywords: Solving Polynomials

Learn more about solving Polynomials at:

#learnwithBrainly

5 0

3 years ago

Read 2 more answers