What were the other new Twelves talking about at school?

What is the vertex of a parabola having the equation y = –2(x – 5)2 + 7? Which direction does the parabola open?

Vika [28.1K]

$y = -2(x-5)^2 + 7$

That's vertex form for a parabola $y=a(x-p)^2+q$ and we read off vertex (p,q) as

Answer: Vertex (5,7)

The negative a tells us this is a downward opening parabola (upside down from the usual $y=x^2$ . I remember CUP - Concave Up Positive; here a is negative so not a cup, instead concave down. The same rule applies to second derivatives in calculus so memorize it now and use it later.

Answer: downward

7 0

3 years ago

The 12th term of a sequence is 41 and the 15th term is 140. What is the nth term?

OLga [1]

An= mth term.
an=a₁+(n-1)*d

a₁₂=41
a₁₅=140

a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d
a₁+11d=41 (1)

a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d
a₁+14d=140 (2)

With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140

we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33

a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322

an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n

an=-355+33n

To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.

3 0

3 years ago

Solve for x. 6x + 5 = -x + 40 B. x= -5 A. x = -9 D. x= 9 C. x = 5

denis23 [38]

C, x=-5, hope this helps

3 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

So there Is the Question Answer This One And the Other is for the other person

ASHA 777 [7]

Answer:

y = 1/3x +5/3

The third one : )

5 0

3 years ago