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lutik1710 [3]
2 years ago
9

HELP!!!!! I do not understand how to do this! Could someone please explain to me how I would solve this problem?

Mathematics
1 answer:
maks197457 [2]2 years ago
3 0
Yeah man nearly there
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Which of the following could be the ratio between the lengths of the two legs of a 30-60-90 triangle check all that apply ....√3
Lerok [7]

Answer:

90 to 30

90/30

Step-by-step explanation:

6 0
3 years ago
If x+y=5,x-y=4,find the value of x²-y²​
inysia [295]

Answer:

20

Step-by-step explanation:

→ First find the value of x and y

x + y = 5

x - y = 4

→ Add the equations to cancel out the y's

2x = 9

→ Divide both sides by 2 to find the value of x

x = 4.5

→ Substitute x = 4.5 back into x - y = 4 to find the value of y

4.5 - y = 4

→ Minus 4.5 from both sides to isolate -y

-y = -0.5

→ Multiply everything by -1

y = 0.5

→ Substitute x = 4.5 and y = 0.5 into x² - y²

4.5² - 0.5² = 20.25 - 0.25 = 20

4 0
3 years ago
What is the y- intercept of the graph of 2x + y = -4?
Brrunno [24]

First write the equation in slope-intercept form which is more commonly known as <em>y = mx + b</em> form where the <em>m </em>or the coefficient of the x term represents the slope of the <em>b</em> or the constant term represents the y-intercept.

Subtract 2x from both sides to get <em>y = -2x - 4</em>.

I put the x term first because that's how it is in y = mx + b form.

Now we can see that the <em>b</em> or the constant term is -4.

We can write this as the ordered pair (0, -4).

Keep in mind when writing a y-intercept as an ordered pair, your x-coordinate will always be 0 in the ordered pair.

8 0
2 years ago
Simplify the expression: 5(x + 2) &gt; 50
kondor19780726 [428]

Answer:

x >8

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
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