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3 years ago

Domain and range of y=6 + -2

Mathematics

1 answer:

g100num [7]3 years ago

6 0

The domain is negative infinity to positive infinity and the range is 8.

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Please help me with this!!!!

olga_2 [115]

The answer would have to be A

5 0

3 years ago

If △NMO ∼ △XYZ, and the area of △NMO is 147 square feet, find the area of △XYZ

sladkih [1.3K]

The area of △XYZ according to the figure is 363ft²

<h3>Area of triangles</h3>

The formula for calculating the area of a triangle is expressed as:

A = 0.5bh

b is the base
h is the height

<h3>Calculate area of each triangles</h3>

For the triangle △NMO

Area of △NMO = 0.5bh

147 = 0.5b(14/12)

147 = 7b/12

7b = 12 * 147

b = 252

b = 252ft

b = 3024in

Get the base of △XYZ

14/3024 = 22/XZ

XZ = 4,752in

Area of △XYZ = 0.5 (4,752)(22)

Area of △XYZ= 52,272in²

Area of △XYZ = 363ft²

Hence the area of △XYZ is 363ft²

Learn more on area of triangles here: brainly.com/question/17335144

6 0

2 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

3 years ago

Read 2 more answers

Don works in home construction. He determined that the function t(A)=144A/9.252 gives the total number of 9 in. tiles needed to

Oksanka [162]

Answer:

6,226 tiles of 9 in are needed to cover an area of 400 square feet

Step-by-step explanation:

Let

A -----> an area in square feet

t(A) ---> the total number of 9 in. tiles needed

we have

$t(A)=\frac{144A}{9.252}$

For A=400 ft^2

substitute in the equation and solve for t

$t(400)=\frac{144(400)}{9.252}$

$t(400)=6,226\ tiles$

That means -----> 6,226 tiles of 9 in are needed to cover an area of 400 square feet

6 0

3 years ago

A restaurant operator in Accra has found out that during the partial lockdown, if she sells a plate of her food for GH¢20 each,

Lilit [14]

Answer:

Step-by-step explanation:

First, note this parameters from the question.

We let x = number of $5 increases and number of 10 decreases in plates sold.

Our Revenue equation is:

R(x) = (300-10x)(10+5x)

We expand the above equation into a quadratic equation by multiplying each bracket:

R(x) = 3000 + 1500x - 3000x - 1500x^2

R(x) = -1500x^2 - 1500x + 3000 (collect like terms)

Next we simplify, by dividing through by -1500

= 1500x^2/1500 - 1500x/1500 + 3000/1500

= X^2 - x + 2

X^2 - x + 2 = 0

Next, we find the axis of symmetry using the formula x = -b/(2*a) where b = 1, a = 1

X = - (-1)/2*1

X = 1/2

Number of $5 increases = $5x1/2 = $2.5

=$2.5 + $20 = $22.5 ticket price gives max revenue.

7 0

3 years ago

Domain and range of y=6 + -2​

Domain and range of y=6 + -2