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natita [175]
1 year ago
13

Is this a function? yes no

Mathematics
1 answer:
zmey [24]1 year ago
3 0
Yes, It’s a function
You might be interested in
Hi what isthe answer
REY [17]

Answer:

2x^7

Step-by-step explanation:

2x^4 × x^3 =

= 2x^(4+3)

= 2x^7

8 0
3 years ago
BRAINLIEST ANSWER!! ASAP<br><br> 3(x+1)+6=-9<br> PLEASE EXPLAIN HOW YOU GOT THE ANSWER
Kryger [21]

Answer:

x=-6

Step-by-step explanation:

use distributive property on the () and times x and 1 by three

then you subtract 6 from both sides

then subtract three from both sides

then divide both sides by 3

your work would look like this:

3(x+1)+6=-9

3x+3+6=-9

3x+3=-15

3x=-18

x=-6

Hope this helps!

8 0
3 years ago
Read 2 more answers
$#4 find the value of x. Round answer to nearest tenth please
Sonja [21]

Answer:

12.9

Step-by-step explanation:

sin 59 = opposite/hypotenuse

opposite is x, the dimension facing the angle 59

hypotenuse is the longest side = 15

sin 59 = x/15

x = 15sin59 = 15 x 0.8572 =12.858 = 12.9 in the nearest tenth

3 0
3 years ago
Simplify.
Ugo [173]


so you take -10.5+5.3+20.2=??

-10.5+5.3=-5.2

-5.2+20.2=15

Ur Answer Is 15

Did I Help??

Hope I Did!!


7 0
3 years ago
Read 2 more answers
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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