Help ASAP math homework

Mathematics

2 answers:

astraxan [27]3 years ago

5 0

$\left\{\begin{array}{ccc}y \ \textgreater \ -x-3\\y\leq4x+2\end{array}\right\\\\y=-x-3\\for\ x=0\to y=-0-3=-3\to(0;-3)\\for\ x=-3\to y=-(-3)-3=3-3-=0\to(-3;0)\\\\y=4x+2\\for\ x=0\to y=4(0)+2=0+2=2\to(0;2)\\for\ x=-1\to y=4(-1)+2=-4+2=-2\to(-1;-2)\\\\Answer\ in\ attachment$

poizon [28]3 years ago

4 0

See attachment for graph.

Notes: graph each line as if they are "equal".
Then draw the lines as:
dashed for < and >
solid for ≤ and ≥
Next, choose a test point and plug it into both equations. If the result is True for BOTH equations, then shade that section.

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Karla invests $300 compounded every 6 months at a rate of 10% for 3 years. At the end of three years, Karla will have $402.03 in

adell [148]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $300

r = 10% = 10/100 = 0.1

n = 2 because it was compounded 2 times in a year(6 months).

t = 3 years

Therefore,

A = 300(1 + 0.1/2)^2 × 3

A = 300(1 + 0.05)^6

A = 300(1.05)^6

A = $402.03

4 0

3 years ago

Suppose each edge of the cube shown in the figure is 10 inches long. Find the sine and cosine of the angle formed by diagonals D

tatiyna

Check the picture below.

$sin(EDG )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{10\sqrt{3}}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{sin(EDG )=\cfrac{\sqrt{3}}{\sqrt{3^2}}\implies sin(EDG )=\cfrac{\sqrt{3}}{3}} \\\\[-0.35em] ~\dotfill$

$cos(EDG )=\cfrac{\stackrel{adjacent}{10\sqrt{2}}}{\underset{hypotenuse}{10\sqrt{3}}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{cos(EDG )=\cfrac{\sqrt{6}}{\sqrt{3^2}}\implies cos(EDG )=\cfrac{\sqrt{6}}{3}}$