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3 years ago

8x - 11 = -2x + 15 -5. What steps would you take to solve this?

1 answer:

8 0

Combine like terms. So that means combine the X's and combine the regular numbers.

- add 2x to 8x
- combine +15 -5
- then add 11 to that
- once you're done, find X by dividing 10x by whatever you have after the = sign

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4 years ago

Someone find area and perimeter of this shape! Thanjs

Gemiola [76]

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6 0

3 years ago

Read 2 more answers

F(x)=-2x^2-2x+10 f(x)=−2x 2 −2x+10 \text{Find }f(2) Find f(2)

romanna [79]

Given that:

f(x) = -2x²-2x+10

f(2) = -2(2)²-2(2)+10

= -2(2*2) - 4 + 10

= -2(4) - 4 + 10

= -8 - 4 + 10

= -12 + 10

= -2 Ans.

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Similar Question

Given f(x) = 1/2x - 5, find f^-1(x) a. f^-1 (x) = -2x+10 b. f^-1(x) = 2x+5 c. f^-1(x) = 2x-10 d. f^-1(x) = 2x+10...

brainly.com/question/11148895?referrer

8 0

3 years ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

5 0

3 years ago

Read 2 more answers

Select the correct answer.

Crazy boy [7]

check the picture below.

now, if we split that triangle like so with that red line, we can use the right-triangle there, to get θ.

$\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{3}}\implies tan(\theta )=1\implies tan^{-1}(\theta )=tan^{-1}(1)\implies \measuredangle \theta =\cfrac{\pi }{4}$

8 0

3 years ago