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3 years ago

15

What is 6x-5>8x+7 but a line under the greater than sign

1 answer:

Ad libitum [116K]3 years ago

7 0

X<-6 with the line under it is the answer, you would flip the sign when u divide

You might be interested in

Write a polynomial function with the real<br> zeros -3, 0, and 3.

suter [353]

$f(x)=x(x-3)(x+3)$

7 0

1 year ago

The function f(x) = –x2 + 20x – 75 models the profit from one customer, in dollars, a shop makes for printing photos, where x is

trasher [3.6K]

The function is graphed as shown below

Part A:

We use the formula $- \frac{b}{2a}$ to find the vertex of the function. A quadratic function of the form of $a x^{2} +bx+c$ and equating this form to the given function $- x^{2} +20x-75$ , we have
$b=20$ and $a=-1$ .

Substituting $a$ and $b$ into the vertex formula, we have
$x=- \frac{20}{2(-1)}=10$ , as shown in the graph

This calculation means that the highest profit is achieved when the number of photo printed equals to ten photos

Part B:

We can find solution to this equation by factorising

$- x^{2} +20x-75=0$
$-( x^{2} -20x+75)=0$
$x^{2} -20x+75=0$
$(x-5)(x-15)=0$
$x=5$ and $x=15$ , as shown in the graph

The two values means that the company makes no profit when they either produce 5 or 15 photos

8 0

3 years ago

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

2 years ago

Ms. Li has 3.56 pounds of grape. She ate 2.1 pounds yesterday. How many pounds of grape are left?

Burka [1]

I think it would have to be 1.46.
Explanation: if you subtract 3.56 from 2.1 you’ll get 1.46

5 0

2 years ago

What is the value of x in the equation 4x+8y=40 when y = 0.8

fgiga [73]

When y=0.8

4x +8(0.8) =40

-> 4x +6.4 =40

-> 4x = 33.4

-> x=33.4/4

-> x=8.35

Hope it helps...

Regards;

Leukonov/Olegion.

8 0

3 years ago