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3 years ago

13

larina recieved a $50 gift card to an online store. she wants to purchase some bracelets that cost $8 each. there will be a $10

overnight delivery fee. solve 8n+10=50 to fine the number of bracelets she can purchase.

2 answers:

BartSMP [9]3 years ago

8 0

N equals 5
hope this helps

aliya0001 [1]3 years ago

7 0

The awnser is:
n=5
So larina can buy 5 bracelets

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6x+2y=8 slope intercept form

77julia77 [94]

Answer:

Step-by-step explanation:

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3 years ago

A triangle with the coordinates X(3, 2), Y(5, -1), and Z(-2, 3) is reflected across both the x- and y-axes. The coordinates afte

nata0808 [166]

Answer:

The coordinates of the reflected triangle are $X'(x,y) = (-3,-2)$ , $Y' (x,y) = (-5,1)$ and $Z'(x,y) = (2,-3)$ .

Step-by-step explanation:

Let be a point $(a,b)\in\mathbb{R}^{2}$ , reflections across x- and y-axes are represented by the following operations:

x-Axis reflection:

$d = d'$

$b - 0 = 0 - b'$

$b' = -b$

y-Axis reflection:

$d = d'$

$a - 0 = 0 - a'$

$a' = -a$

If we know that $X (x,y) = (3,2)$ , $Y(x,y) = (5,-1)$ and $Z (x,y) = (-2, 3)$ , the coordinates after both reflections are, respectively:

$X'(x,y) = (-3,-2)$

$Y' (x,y) = (-5,1)$

$Z'(x,y) = (2,-3)$

The coordinates of the reflected triangle are $X'(x,y) = (-3,-2)$ , $Y' (x,y) = (-5,1)$ and $Z'(x,y) = (2,-3)$ .

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4 years ago

In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the

natka813 [3]

Answer:

$P(t) = 5100e^{0.0039t}$

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that $P(0) = 5100$

So

$P(t) = 5100e^{rt}$

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that $P(5) = 5200$ .

We use this to find r.

$P(t) = 5100e^{rt}$

$5200 = 5100e^{5r}$

$e^{5r} = \frac{52}{51}$

$\ln{e^{5r}} = \ln{\frac{52}{51}}$

$5r = \ln{\frac{52}{51}}$

$r = \frac{\ln{\frac{52}{51}}}{5}$

$r = 0.0039$

So the equation for the moose population is:

$P(t) = 5100e^{0.0039t}$

5 0

3 years ago

Hi ! please help !

tatuchka [14]

Hey there! I was looking at your question for a bit then realized you meant $i$ , the imaginary number, not "1" at the end of each number.

Complex numbers are numbers involving both real and imaginary numbers; thus, this question wouldn't make much sense without the imaginary number $i$

The expression we are given is:

$(3+4i)+(8+2i)$

Open parenthesis and combine like terms:

$=3+4i+8+2i$

$=11+6i$

This is answer choice B (assuming it's supposed to be an $i$ and not a "1" at the end)

Let me know if you need any clarifications, thanks!

~ Padoru

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4 years ago

Read 2 more answers

Pls help me help me pls

grin007 [14]

Answer:

Point B

Step-by-step explanation:

Lines given:

y= 2/5x - 1/2
y= -1/3x +2/3

x coordinate of intersection of given lines is:

2/5x - 1/2= -1/3x+2/3
2/5x+1/3x= 2/3 +1/2
6/15x+5/15x= 4/6+3/6
11/15x=7/6
x= 7*15/11*6
x= 35/22

For this value of x, the value of y is:

y= 2/5*35/22 - 1/2
y=14/22- 1/2
y= 3/22

As per graph point (35/22, 3/22) is point B

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3 years ago