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3 years ago

NEED HELP ASAP!!! Giving brainliest!!!!!!!

Mathematics

2 answers:

KengaRu [80]3 years ago

5 0

The answer is c I agree with the answers above

attashe74 [19]3 years ago

3 0

C.(f-g)(x) = 4x^3 +5x²-7x-1

Step-by-step explanation:

Given information :

$f(x) = 4 {x}^{3} + 5 {x}^{2} - 3x - 6 \\ g(x) = 4x - 5$

Find :

$(f - g)(x) = \\ (4 {x}^{3} + 5 {x}^{2} - 3x - 6) \\ - 4x -5$

Open bracket and Simplify

$4 {x}^{3} + 5 {x}^{2} - 3x - 6 - 4x + 5 \\ 4 {x}^{3} + 5 {x}^{2} - 7x - 1$

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Using the quadratic formula to solve x2 + 20 = 2x, what are the values of x??

azamat

$x^2-2x+20=0\implies D=b^2-4ac$ .

There are no real solutions to the quadratic equation.

Hope this helps.

6 0

3 years ago

MATHEMATICS

Andrej [43]

Second one the real answer is

<span>0.219512195122</span>

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3 years ago

Read 2 more answers

Which expression is equivalent to log Subscript w Baseline StartFraction (x squared minus 6) Superscript 4 Baseline Over RootInd

lora16 [44]

The logarithmic expression $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ can be expressed as $\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$ .

Given to us

$\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$

<h3>Which expression is equivalent to $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ ?</h3>

To solve the problem we will use the basic logarithmic properties,

$\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$

Using the logarithmic property $\rm log_a\dfrac{x}{y} = log_ax-log_ay$ ,

$=\rm log_w{(x^2-6)^4} - log_w{\sqrt[3]{x^2-8}}$

Using the exponential property $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ ,

$=\rm log_w{(x^2-6)^4} - log_w(x^2-8)^{\dfrac{1}{3}}$

Using the logarithmic property $\rm log_aB^x = xlog_aB$ ,

$=\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$

Hence, the logarithmic expression $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ can be expressed as $\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$ .

Learn more about Logarithmic Expression:

brainly.com/question/7302008

5 0

3 years ago

In Mary's first math test she scored 87%. The mean and standard deviation for the class were 71% and 18% respectively. In her se

erica [24]

Answer:

Step-by-step explanation:

Since your population are the students in math class, you can use the z-score formula $z=(x-\mu)/\sigma$ in order to comparing the two math test scores. Where $\mu$ is the mean for the class, $\sigma$ is the standars deviation and x is Mary score.

For the first test $\mu=.71 , \sigma=.18,x=.87$ , so , $z_{1} = (.87-.71)/(.18)=.88$ .

For the second test $\mu=.53 , \sigma=.14,x=.66$ , so , $z_{1} = (.66-.53)/(.14)=.93$

Mary do better in the second test, relative to the rest of the class (because $.88 \leq .93$ , it means the second score is nearer to the mean score of the class than the first one )