HI. I really Need help. I will Mark Brainliest!

Answer:

a=4

Step-by-step explanation:

8=3a-4

12=3a

a=4

Answer:

you add 40+45

Step-by-step explanation:

40+45=85

then you subtract 180 - 85 = 95

then divide x into 95 that will give you x=95

Answer:

[tex]x > \frac{1}{4} [/tex]

Step-by-step explanation:

[tex]12x - 5 > - 2[/tex]

Lets add +5 in both the sides.

[tex] = > 12x - 5 + 5 > - 2 + 5[/tex]

[tex] = > 12x > 3[/tex]

Now lets divide 12 on both the sides.

[tex] = > \frac{12x}{12} > \frac{3}{12} [/tex]

[tex] = > x > \frac{1}{4} [/tex]

Answer:

hard to read, can you take another picture please?

Step-by-step explanation:

Answer:ugh is there an picture maybe i can understand it better?

Step-by-step explanation:

Answer:

d 5 tan theta

Step-by-step explanation:

5 sin theta x 1/cos theta

5 (sin theta/cos theta)

5 tan theta

Answer:

can you show us the graph

Answer:

t=0.08

Step-by-step explanation:

see attached kindly

Answer:

13

Step-by-step explanation:

Given data

flat amount =$10

let the number of dogs walked be x

and let the money earned be y

y= 10+5x

for y= 75

let us find x

75= 10+5x

75-10= 5x

65= 5x

divide both sides by 5

x= 65/5

x=13

Hence the number of dogs is 13

Hey there!

Step By Step:

Well so we know that 12 is about half of an 25. So what we are gonna do is to buy 3 of 12$ and 2 of 25$.

3x12=36 and 2x25=50.

With adding those two last numbers, you’ll get $86.

Graph would be down below.

Hope it helps :D

Answer:

-7

Step-by-step explanation:

Answer:

la eneria es muy importante

Answer:

x=7

Step-by-step explanation:

∠STP=∠SWT+∠TSW(exterrior ∠ of Δ)

15x=7x+1+7x+6

x=7

Answer:

12 because u use the other shapes to reference your measurements.

Answer:

15

Step-by-step explanation:

8^2 + b^2 = 17^2

64 + b^2 = 289

289 - 64 = 225

square root of 225 = 15

Answer:

3√3−x=3√2x

Graph each side of the equation. The solution is the x-value of the point of intersection.

x≈0.46567646

image of graph

Step-by-step explanation:

Hope it is helpful.....

The fractional form is 7/10.

The fractional bar is a horizontal bar that divides the numerator and denominator of every fraction into these two halves.

Given:

We have to divide 7 by 10.

The common factor for 7 and 10

7 = 7 x 1

and, 10= 2 x 5 x 1

so, there is only one factor common is 1.

Thus, the fraction is 7/10.

Learn more about Equivalent Fraction here:

https://brainly.com/question/29766013

#SPJ2

Answer:

eight hundred seventy-five thousand six hundred eighty-seven

hope this helps :)

Answer:

A. {–8, –2, 18}

Step-by-step explanation:

y = 5x − 2

x = −1.2 ----> y = 5(−1.2) − 2 = −6 − 2 = −8

x = 0 -----> y = 5(0) − 2 = 0 − 2 = −2

x = 4 -----> y = 5(4) − 2 = 20 − 2 = 18

Answer:

x = 11

Step-by-step explanation:

3(11 - 10) + 7 =

3 + 7 = 10

hope this helped

Answer:

Explanation:

(-6 + 3i) + (6 + 4i)

-6 + 6 = 0

3i + 4i = 7i

Answer:

x = 3

Step-by-step explanation:

Answer: x = 3

Step-by-step explanation:

Answer:

i cant help you with this sorry

Step-by-step explanation:

Answer:

[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1[/tex]

Step-by-step explanation:

We are given:

[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)[/tex]

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

[tex]=(1-1)^\ln(1)}=0^0[/tex]

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

[tex]y=(x-1)^\ln(x)[/tex]

By taking the natural log of both sides:

[tex]\ln(y)=\ln(x)\ln(x-1)[/tex]

And by taking the limit as x approaches 1 from the right of both functions:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)[/tex]

Rewrite:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}[/tex]

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}[/tex]

Simplify:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)[/tex]

Simplify:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}[/tex]

Now, by using direct substitution, we will acquire:

[tex]\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}[/tex]

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))[/tex]

Finally, direct substitution yields:

[tex]\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0[/tex]

Thus:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=0[/tex]

By the Composite Function Property for limits:

[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0[/tex]

Raising both sides to e produces:

[tex]\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0[/tex]

Therefore:

[tex]\displaystyle \lim_{x\to 1^+}y=1[/tex]

Substitution:

[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1[/tex]