What is x squared 6x - 7 factorised?

Seth and Karen are college students who take on different jobs for income. Seth needs $750 for expenses and savings each month;

pantera1 [17]

Answer:

Seth:

Equation 7.50x + 15y = 750, where x is the number of hours Seth works and y is the number of lawns he mows.

Karen:

Equation 6x + 4y = 700, where x is the number of hours Karen works and y is the number of dogs she walks.

Part A: Converting it in function for by solving for y first.

Solving 7.50x + 15y = 750, equation we get

7.50x + 15y = 750

15y = -7.50x + 750

y = -1/2x + 50

Solving 6x + 4y = 700 for y, we get

6x + 4y = 700

4y = -6x + 700

y= -3/2x + 175.

On comaring with slope-intercept form y=mx+b, we got slope for Seth equation is -1/2 and slope for Karen is -3/2.

If we take absolute of those -1/2 and -3/2 we get 1/2 and 3/2.

Therefore, Karen has the higher slope. We can see the graph that blue line has greater slope as it's increasing/decreasing with greater rate.

In function form we could write equation as

f(x) = -1/2x + 50 and

f(x) = -3/2x + 175.

Part B : y-intercept of Karen equation is 175 and y-intercept of Seth equation is 50.

Therefore, Karen has greater y-intercept.

This means Karen earns $175 by just walking dogs.

Part C: If both students work 80 hours for the month.

Let us plug x =80 in each of the equations.

y = -1/2x + 50 => y = -1/2(80) + 50 = -40 +50 = $10.

Seth would earn $10 from mowing lawn.

Therefore, the number of lawns Seth has to mow = 10/15 = 0.67 that is approximately 1 lawn.

y = -3/2x + 175 => y = -3/2(80) + 175 = -120 +175 = $55.

Karen would earn $50 from walking dog.

So, the number of dogs Karne has to walk = 55/4 = 13.75 that ia approximately 14 dogs.

Therefore, the number of lawns Seth has to mow is less than the number of dogs Karen has to walk.

Part D: If Karen begins charging $8 per hour for babysitting and Seth begins earning $8 per hour.

The equations would become : 8x + 15y = 750, 8x + 4y = 700.

It would not effect the graph much. Even the y-intercepts would remain same.

There would be a slightly change in slopes.

Step-by-step explanation:

8 0

3 years ago

When you divide a fraction less than one by a whole number greater than one, is the Quotient less than, greater than, or equal t

Olin [163]

Answer:

bw eghw d32h n wqkjrd k2wsdtep explanation: less than ewrdi2qwed

5 0

3 years ago

Only problem 6a. Can someone demonstrate how to do this so I can do the rest solo?

Lisa [10]

as far as I can tell, is simply asking to write two more expressions, that are equivalent to the provided one, namely, grab the provided one and expand it, if you simplify the expanded version, you'd end up with the provided, for example

$\bf \boxed{6a.1}~\hfill \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}}\\\\\\\boxed{6a.3}~\hfill \stackrel{ch anging}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}$

so let's do 6a1, 6a3 and 6a5.

$\bf \boxed{6a.1}~\hfill \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}} \\\\\\ \boxed{6a.3}~\hfill \stackrel{changing}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}$

$\bf \boxed{6a.5}~\hfill \stackrel{changing}{(9\cdot 10)\div (2\cdot 10)}\implies \stackrel{one}{90\div 20}~\hfill \stackrel{changing}{(9\cdot 70)\div(2\cdot 70)}\implies \stackrel{two}{630\div 140}$

3 0

3 years ago

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and

inn [45]

Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)

t=8.169

P-value= ?

Answer:

a) $P-value=0$