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nikitadnepr [17]
3 years ago
10

Graph 5x+y=2 and 5x-y=4

Mathematics
1 answer:
Andru [333]3 years ago
7 0

Answer:

What do you mean

Step-by-step explanation:


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Pls help me with this​
Free_Kalibri [48]

Answer:

\boxed{\sf \frac{15}{22} }

Step-by-step explanation:

The radius of the circle is 7 cm.

The two legs of the triangles are 7cm as well.

Area of a trinagle is ( base × height )/2.

\frac{7 \times 7}{2} =24.5

Multiply the value by 2 since there are two triangles.

24.5 \times 2 = 49

Calculate the area of the circle.

\pi r^2

The radius is given.

\pi (7)^2

Take \pi as \frac{22}{7}

49(\frac{22}{7} )

=154

Subtract the area of the two triangles from the area of the whole circle.

154-49=105

The area of the shaded part is 105 cm². The total area of the shape is 154cm².

\frac{105}{154} =\frac{15}{22}

5 0
3 years ago
What is the reciprocal of 2/3​
galina1969 [7]

Answer:

3/2 or 1 and a half

hope it helps.

just sitting here waiting for brainliest........

7 0
3 years ago
Please help me pleaseeee
GenaCL600 [577]

Answer:

∡PQS = 35+14=49

Step-by-step explanation:

4 0
3 years ago
Doug had about $3,215 in savings while his brother had double that amount in savings. What is a reasonable amount that the broth
MaRussiya [10]

Answer: i think you would have to multiply

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
3 years ago
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