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Nuetrik [128]
4 years ago
10

Alx+b) = 4x + 10

Mathematics
1 answer:
Leokris [45]4 years ago
3 0

Answer:

a=4, b=5/2

Step-by-step explanation:

Assuming the given equation is

a(x + b) = 4x + 10

We expand the left hand side to get:

ax + ab = 4x + 10

If this equation must have infinitely many solutions, then

ax = 4x \\  \implies \: a = 4

Also we must have

ab = 10

Put a=4 to get:

4b = 10

b =  \frac{10}{4}  =  \frac{5}{2}

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Draw a model to show a fraction that is equivalent to 1 thirds and a fraction that is not equivalent to 1 thirds.
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A fraction equal to 1/3 is 2/6 so you can draw six circles, and color in two of them

4 0
3 years ago
Simplify the expression.<br> 1) √65/121<br> 2) √25/36
kakasveta [241]
√65/√121=√65/11
√25/√36=5/6
I am not sure if that is your problem are the denominators square root too if so u just need to reduce and simplify. √65 is not a perfect square therefore you cannot reduce it cause it does not contain any perfect squares in its factors either. 
6 0
4 years ago
Evaluate the limits<br><br>​
julsineya [31]

x > \ln(x) for all x, so

\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}

Similarly, \displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}

We can of course see the limits are identical by replacing x\mapsto e^x, so that

\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)

Using the rule, the limit here is

\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'}  = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0

so the overall limit is

\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty

6 0
2 years ago
Help. I just started this​
inn [45]
X is (4/3,0)
Y is (0,-4/5)


Hope that helps!
7 0
3 years ago
Please help ill mark brainliest​
lesantik [10]

Answer:

I think its A ratio reduced, since a scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure.

7 0
3 years ago
Read 2 more answers
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