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3 years ago

3.) T4.T4 Simply this for me to plz look at the pic

Mathematics

1 answer:

nikdorinn [45]3 years ago

4 0

Answer:

t^8

Step-by-step explanation:

t^4 times t^4

exponents you add.

4 + 4 = 8

t^8

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Determine whether the function has a maximum or minimum value. f(×)= x^2+9

PtichkaEL [24]

$f(x)= x^2+9\\\\for\ each\ x\in R\\\\x^2 \geq 0\ \ \ \Rightarrow\ \ \ x^2+9 \geq 9\ \ \ \Rightarrow\ \ \ f(x) \geq 9\\\\y_{min}=9\\ y_{max}\ \ \ \rightarrow\ \ \ not\ exist$

3 0

4 years ago

Read 2 more answers

Rewrite the equation in vertex form. then find the vertex of the graph. y=-3x^2-5x+1

amm1812

Answer: y = -3(x + $\frac{5}{6}$ )² + $\frac{37}{12}$ , $(-\frac{5}{6}$ , $\frac{37}{12})$

Step-by-step explanation:

First, you need to complete the square:

y = -3x² - 5x + 1

-1 -1

y - 1 = -3x² - 5x

y - 1 = -3(x² + $\frac{5}{3}x$

y - 1 + -3( $\frac{25}{36}$ ) = -3(x² + $\frac{5}{3}x$ + $\frac{25}{36}$ )

↑ ↓ ↑

$\frac{5}{3*2}$ = $(\frac{5}{3*2})^{2}$

y - 1 - $\frac{25}{12}$ = -3(x + $\frac{5}{6}$ )²

y - $\frac{12}{12}$ - $\frac{25}{12}$ = -3(x + $\frac{5}{6}$ )²

y - $\frac{37}{12}$ = -3(x + $\frac{5}{6}$ )²

y = -3(x + $\frac{5}{6}$ )² + $\frac{37}{12}$

Now, it is in the form of y = a(x - h)² + k where (h, k) is the vertex

Vertex = $(-\frac{5}{6}$ , $\frac{37}{12})$

8 0

3 years ago

High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capa

zalisa [80]

Answer:

The students have to score 0.74 standard deviations above the mean to be publicly recognized.

Step-by-step explanation:

A random variable X is said to have a normal distribution with parameters µ (mean) and σ² (variance).

If $X \sim N (\mu, \sigma^{2})$ , then $z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

The distribution of these z-scores is known as the standard normal distribution.

The z-score is a standardized form of the raw score, X. It is a numerical measurement of the relationship between a value (X) and the mean (µ) in terms of the standard deviation (σ). A z-score of -1 implies that the data value is 1 standard deviation below the mean. And a z-score of 1 implies that the data value is 1 standard deviation above the mean.

Let X be defined as the scores of students at the National Financial Capability Challenge Exam.

It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.

That is, P (X > x) = 0.23.

⇒ $P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma})=0.23$