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3 years ago

F() = x2 + 1 g(x) = 5 - x (f+g)(x) =

Mathematics

1 answer:

frutty [35]3 years ago

4 0

Answer:

(f+g)(x) = x² - x + 6

Step-by-step explanation:

We can find (f+g)(x) by adding f(x) and g(x).

f(x) = x² + 1

g(x) = 5 - x

(f+g)(x) = f(x) + g(x)

(f+g)(x) = (x² + 1) + (5 - x)

(f+g)(x) = x² + 1 + 5 - x

(f+g)(x) = x² - x + 6

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A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square roo

deff fn [24]

Answer:

$\frac{x^2}{61}-\frac{y^2}{37} =1$

Step-by-step explanation:

The standard equation of a hyperbola is given by:

$\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1$

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

$\frac{x^2}{61}-\frac{y^2}{37} =1$

6 0

3 years ago

HELP PLEASE <br> solve the right triangle

aleksandrvk [35]

Answer:

Part 1) $FG=4.4\ units$

Part 2) $EF=3.9\ units$

Part 3) $m\angle G=63^o$

Step-by-step explanation:

step 1

Find the measure of length side FG

In the right triangle EFG

we know that

$sin(27^o)=\frac{GE}{FG}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(27^o)=\frac{2}{FG}$

$FG=\frac{2}{sin(27^o)}=4.4\ units$

step 2

Find the measure of length side EF

In the right triangle EFG

we know that

$cos(27^o)=\frac{EF}{FG}$ ----> by CAH (adjacent side divided by the hypotenuse)

substitute the given values

$cos(27^o)=\frac{EF}{4.4}$

$EF=cos(27^o)(4.4)=3.9\ units$

step 3

Find the measure of angle G

we know that

$m\angle G+27^o=90^o$ ---> by complementary angles in a right triangle

$m\angle G=90^o-27^o=63^o$

7 0

3 years ago

I need help with this <br> ASAP!!!!!!!!!

zzz [600]

Answer:

<e & <g are vertical

<e & <a are corresponding

<e & <a are alternate interior angles

<g & <a are alternate exterior angles

<g & <f are none of the above

ur welcome. have a great day

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3 years ago

What are the zeros of the function f(x)= 2x^2-4x-6

jolli1 [7]

The root are x=3 and x=-1

youre welcome :)

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3 years ago

How do you solve y=3x+16 using substitution

Ghella [55]

There's nothing to substitute, and no solution,
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3 years ago