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3 years ago

6

If a function is defined by the formula y equals 2x + 1 + that domain is given by the set 3, 5, 7, 9 then which set represents t

he range of function

1 answer:

Mariulka [41]3 years ago

8 0

Answer:

{7, 11, 15, 19}

Step-by-step explanation:

The function is defined by the formula y = 2x + 1 ........ (1)

Now, domain of the function are given to be {3, 5, 7, 9}

Hence, from equation (1), at x= 3, y = 2 × 3 + 1 = 7

Now, at x = 5, y = 2 × 5 + 1 = 11

Again, at x = 7, y = 2 × 7 + 1 = 15

Finally, at x = 9, y = 2 × 9 + 1 = 19

Therefore, the range of the corresponding domain of this function is given to be {7, 11, 15, 19}. (Answer)

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I need help on how to solve this so anything is good :)

DanielleElmas [232]

Answer:

STUV is a square

Step-by-step explanation:

segment length² = (x-x₁)² + (y-y₁)²

ST²: (-9 - 1)2 + (14 - 10)² = (-10)² + 4² = 116 (the rest follow this formula)

TU² = 116 TV² = 232 SU² = 232 SV² = 116 UV² = 116

ST = TU =SV = UV (4 sides congruent)

TV = SU (diagonal equal)

This is a square

4 0

2 years ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

3 years ago

Read 2 more answers

Add: <br> 3x+9+8x + 12<br> 2x + 6+x² + 6x + 9

Oxana [17]

I'll treat these like they're two seperate problems because how you set it up they're not together. If it's one whole problem please tell me and I'll solve it that way. :-)

First one: 3x+9+8x + 12

Collect like terms and simplify

(3x+8x)+(9+12)

11x+21

Second one: 2x + 6+x² + 6x + 9

Collect like terms and simplify

$(2x+6x)+(6+9)+x^{2}$

8x+15+ $x^{2}$

Hope this helps you, have a BLESSED and wonderful day, as well as a safe one!

-Cutiepatutie ☺❀❤

6 0

3 years ago

Don’t understand could I get some help pls :)

Lemur [1.5K]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Parallel lines.

We must first understand that, Parallel Lines always have a same Slope, hence the 'm' value in y=mx+c equation will same, here it is '1/2' in the above equation,

so the points here are (-6,-17)

==>

(y-(-17)) = m(x-(-6))

==>

here m = 1/2 ,hence

y+17 = 1/2(x+6)

==> y+17 = 1/2(x) + 3

==> y = 1/2(x) + 3 - 17

==> y = 1/2(x) - 14

hence the Option 4.) is the correct answer!!

5 0

3 years ago

A dragonfly can beat its wings 30 times a second. write an equation in slioe intercept form that shows the relationship between

ivanzaharov [21]

Answer:

1800

Step-by-step explanation:

Since there is the tragon fly beat his wings 30 times a second ,there is 60 hundredths of a second in a second multiply 30 x 60 and you should get 1800.

mulitiply 0x0 = 0

2. mulitiply 6x0=0

3. multiply 3x0=0

4. mulitiply 6x3 = 18

5. have 1800

or you can just multiply 6x3 = 18 and add the two zeros

6 0

3 years ago