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Viefleur [7K]
3 years ago
8

I WILL GIVE BRAINLIEST Solve for t 4 (t + 1/4) = 3 t =

Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0

Answer:

t= 1/2

Step-by-step explanation:

hope this help

Katyanochek1 [597]3 years ago
7 0

The answer is t = 1/2.

Have a nice day to whoever is reading this! :D

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Use synthetic division to find (5x<br> + 31x + 32) = (x + 5).
LiRa [457]
We move all terms to the left:
+31x+32)-((+5)=0
So your answer is 0
5 0
3 years ago
Which answer is the best estimate of the correlation coefficient for the variables in the scatter plot?
Vesnalui [34]
Its Letter B -0.5

negative because its slope is negative (from left to right it's going downward)

so there are 2 negavite numbers in the choices, -0.5 and -0.95.

The nearer the value to 1 the less scatter it is. The farther the value to 1 (means nearer to 0) the more scatter it is. Because the plot in the picture is mostly scattered the value should be nearer to 0. It is negative 0.5.
6 0
3 years ago
Read 2 more answers
Is a relation always a function? Is a function always a relation? Explain.
katen-ka-za [31]

A function is always a relation but relation is not always a function

<u>Solution:</u>

Given that, we have to explain Is a relation always a function and is a function always a relation

Note that both functions and relations are defined as sets of lists.  

In fact, every function is a relation. However, not every relation is a function.  A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y.

That is, given an element x in X, there is only one element in Y that x is related to.

For example, consider the following sets X and Y. Let me give you a relation between them that is not a function;

X = { 1, 2, 3 }

Y = { a , b , c, d }

Relation from X to Y : { (1,a) , (2, b) , (2, c) , (3, d) }

This relation is not a function from X to Y because the element 2 in X is related to two different elements, b and c

Relation from X to Y that is a function: { (1,d) , (2,d) , (3, a) }

This is a function since each element from X is related to only one element in Y. Note that it is okay for two different elements in X to be related to the same element in Y. It's still a function, it's just not a one-to-one function.

So, we can say that function is a type of relation.

Which means whatever a function occurs, it will be a relation from one set to other.

But when a relation occurs it may be a function but need not be always a function.

Hence, a function is always a relation but relation is not always a function.

8 0
3 years ago
HELP!!! What are the quotient and remainder when x^3 + 9x^2 + 21x + 24 is divided by x + 2? PLEASE HELP
Sedaia [141]

Answer:

A) The quotient is x^2+7x+7 and the remainder is 10.

Step-by-step explanation:

Set up the synthetic division using the coefficients of the numerator and the root in the denominator. Divide using the rules for synthetic division.

4 0
3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
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