PLEASE HELP AND SHOW YOUR WORK IM CONFUSED!

For what values of m does the graph of y = 3x2 + 7x + m have two x-intercepts?

frutty [35]

Answer:

m < $\frac{49}{12}$

Step-by-step explanation:

Using the discriminant Δ = b² - 4ac

Given a quadratic equation in standard form, y = ax² + bx + c

Then the value of the discriminant determines the nature of the roots

• For 2 real roots then b² - 4ac > 0

Given

y = 3x² + 7x + m ← in standard form

with a = 3, b = 7 and c = m, then

7² - (4 × 3 × m) > 0

49 - 12m > 0 ( subtract 49 from both sides )

- 12m > - 49

Divide both sides by - 12, reversing the symbol as a result of dividing by a negative quantity.

m < $\frac{49}{12}$

8 0

3 years ago

Read 2 more answers

I need the answer to these, to be right, and you got to show the work on how you got the answer because if not I’ll get them wro

alexira [117]

A: $\frac{5}{3} = \frac{N}{12}$
Multiply both sides by 12
Answer: $N = 20$

B: $\frac{7}{2} = \frac{42}{N}$
Multiply both sides by 2, $7 = \frac{84}{N}$
Multiply both sides by N, $7N = 84$
Divide both sides by 7,
Answer: $N=12$

C: $\frac{N}{2}= \frac{50}{N}$
Multiply both sides by 2, $N = \frac{100}{N}$
Multiply both sides by N, $N^{2} = 100$
Simplify,
Answer: $N = 10 , -10$

Hope I helped you!

4 0

3 years ago

Read 2 more answers

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty

Sophie [7]

Answer:

The interval is $0.7187 < p < 2.421$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 100$

The population proportion is $p = 0.81$

The confidence level is C = 98%

The level of significance is mathematically evaluated as

$\alpha = 100 -98$

$\alpha = 2%$ %

$\alpha = 0.02$

Here this level of significance represented the left and the right tail

The degree of freedom is evaluated as

$df = n-1$

substituting value

$df = 100 - 1$

$df = 99$

Since we require the critical value of one tail in order to evaluate the 98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The critical value of $\frac{\alpha}{2}$ and the evaluated degree of freedom is

$t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33$

this is obtained from the critical value table

The standard error is mathematically evaluated as

$SE = \sqrt{\frac{p(1-p )}{n} }$

substituting value

$SE = \sqrt{\frac{0.81(1-0.81 )}{100} }$

$SE = 0.0392$

The 98% confidence interval is evaluated as

$p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }$

substituting value

$0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392$

$0.7187 < p < 2.421$

4 0

3 years ago

By rounding each number to 1 significant figure,

luda_lava [24]

Answer:

Step-by-step explanation:

1. 8.2 * 6.7 / 0.46

2. 54.94 / 0.46

3. 119.434783

4: 120

4 0

3 years ago

Out of the 400 students in a final year in secondary school, 300 are offering biology and 190 are offering chemistry. I) how man

iren2701 [21]

Answer:

I) how many student are offering both subject , if only 70 students are offering neither biology nor chemistry?

= 160 students

II) how many students are offering at least one biology and chemistry?

= 330 students

Step-by-step explanation:

I) how many student are offering both subject , if only 70 students are offering neither biology nor chemistry?

Total number of student n ( B ∪ C) = 400

Students offering biology n(B) = 300

Students offering chemistry n(C) = 190

Students not offering biology nor chemistry = 70

Students offering both biology and chemistry n ( B ∩ C) = ??

n ( B ∪ C) - Students neither offering biology nor chemistry = n(B ) + n ( C ) - n ( B ∩ C)

400 - 70 = 300 + 190 - n ( B ∩ C)

330 = 490 - n ( B ∩ C)

n ( B ∩ C) = 490 - 330

n ( B ∩ C) = 160

Therefore, students offering both Biology and Chemistry are 160 students.

II) how many students are offering at least one biology and chemistry?

n(B' ) + n ( C') + n ( B ∩ C)

n(B') = Students offering biology only

= n(B) - n ( B ∩ C)

= 300 - 160 = 140

n(C') = Students offering chemistry only

n(C) - n ( B ∩ C)

= 190 - 160

= 30

(300 - 160) + (190 - 160) + 160

140 + 30 + 160

= 330 students

4 0

3 years ago