Write 6 × 10 -7 in decimal notation. 0.0000006 0.00006 600,000 6,000,000

Which inequality has no solution? 6(x+2) > x-3 3+4x2(1+ 2x) -2(x+6) X-9 <3(x-3)

Juli2301 [7.4K]

Answer:

$6(x + 2) > x - 3$

$6x + 12 > x - 3$

$6x - x > - 12 - 3$

$5x > - 9$

$3 + 4 \times 2(1 + 2x)$

$3 + 8(1 + 2x)$

$3 + 8 + 16x$

$11 + 16x$

$- 2(x + 6)x - 9 < 3(x - 3)$

$- 2( {x}^{2} + 6x) - 9 < 3x - 9$

$- 2 {x}^{2} - 12x - 9 < 3x - 9$

5 0

2 years ago

The table shows the ratio of caramel corn to cheddar corn in a bag. What is the constant of proportionality?

Anit [1.1K]

For proportionality constant problems, set up the equation as,

$x=Ky$ ,

Where x and y are the two variables you are comparing and K is the proportionality constant. If we take Caramel Corn values as x and Cheddar Corn values as y, and then solve for K for each ratio lines, we will get the same answer. Let's check.

$15=K(10)\\K=\frac{15}{10}= \frac{3}{2}$ ,

$30=K(20)\\K=\frac{30}{20}= \frac{3}{2}$ , and

$45=K(30)\\K=\frac{45}{30}= \frac{3}{2}$ .

Hence, the proportionality constant, in this case K, is equal to $\frac{3}{2}$ or 1.5. First answer choice is correct.

ANSWER: 1.5

4 0

2 years ago

Read 2 more answers

Consider a manufacturing process with a quality inspection station. In the past, 10% of parts are defective. As soon as one defe

Kruka [31]

Answer:

0.9

Step-by-step explanation:

10% is equal to 0.1

The probability of having defective parts in a pile of parts is 0.1

Before the process is stopped, 1 part has to be defective.

In a pile of 9 parts, the probability that a part is defective 0.1 of 9, which is = 0.9 hence, approximately one (1) part will be defective in a pile of 9 parts and the process will be stopped.

Since there was no defective part among the first 6 parts, P(d) was 0

That is, probability of a defective part was zero.

7 0

2 years ago

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Alla [95]

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

5 0

2 years ago

Please help!!! Multiply. Write your answer in scientific notation. (5×10 – 8)(1×103) 20 pts

PolarNik [594]

Answer:

4.326 x 10^ 3

Step-by-step explanation:

Correct me if wrong please :)

4 0

2 years ago