All categories

3 years ago

I need help i can't get the right answer

Mathematics

2 answers:

ziro4ka [17]3 years ago

8 0

3x-26=2x+24

3x-2x=24+26

x=50

please mark this answer as brainlist

dimulka [17.4K]3 years ago

4 0

Answer:

x = 50

Step-by-step explanation:

I guess, this diagram tells us that the lines are parallel.

that means that they cut the third line at the exact same angles.

and that means

3x - 26 = 2x + 24

x - 26 = 24

x = 50

You might be interested in

How do I find the difference in the simplest form?

GrogVix [38]

Answer:

Step-by-step explanation:

$\frac{8x}{4x - 7} - \frac{14}{4x - 7} \\ \\ = \frac{8x - 14}{4x - 7} \\ \\ = \frac{2 \cancel{(4x - 7)}}{\cancel{(4x - 7)}} \\ \\ = 2$

5 0

2 years ago

Beginning in 1996,a glacier lost an average of 3.7 meters of thickness each year.Find the total change in it's thickness by the

Elza [17]

Take the common denominator and divide by 3 then youll get 58

8 0

4 years ago

Factor 15x^2+20x that's my question

liq [111]

Factor out the GCF
15x^2+20x

15x^2/5x= 3x
20x/5x= 4

Final answer: 5x(3x+4)

8 0

4 years ago

PLEASE HELPP!!!

Andrew [12]

Answer:

It would look like this basically..

Step-by-step explanation:

If a function, f(x) is shifted to the left four units, the

function that represents the transformation would be f(x+4). Adding it would make it go right..

*If you ever need help graphing, use desmos!

4 0

4 years ago

A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was chec

PSYCHO15rus [73]

Answer:

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was $\bar{x} = 686.75$ with a standard deviation of s = 256.20. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the 100 $(\alpha/2)$ th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a $100(1-\alpha)$ % = 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $686.75\pm t_{0.01}(\frac{256.20}{\sqrt{15}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 14 df, i.e., $t_{0.01} = -2.6245$ . Then, we have $686.75\pm (-2.6245)(\frac{256.20}{\sqrt{15}})$ and the 98% confidence interval is given by (513.1379, 860.3621)

7 0

4 years ago