All categories

3 years ago

7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and

Mathematics

1 answer:

Juliette [100K]3 years ago

3 0

Answer:

A) JK and AB =

$[tex] \boxed{( \theta) = \f{12}{( \theta)} }$ So, =》 $\dfrac{10}{5} = \dfrac{1}{ \( \theta) }$ =》 $\( \theta) = \dfrac{5}{2}$ [/tex]

B) AABC

$[tex] \boxed{( \theta) = {1}{( \theta)} }$ So, $23}{-18=\\Ans } \boxed$

C) AABM:

$[tex] e \fbox{: }$ The value of k for which $\sin(jm) = \cos(x)$ is $\dfrac{\p}{2}$ ans[/tex]

----------

You might be interested in

Hello! This is my first question ever. I am in AP Calculus and am working on a redo. I know the correct answer but don't know ho

Kipish [7]

Since $\Delta x=\dfrac{b-a}n$ , and you have a corresponding term in the given Riemann sum of $\dfrac5n$ , you know the integral is being taken over an interval of length 5, so you can omit the second choice.

Next, $f(a+i\Delta x)$ corresponds to $a=3$ with $\Delta x=\dfrac5n$ . The fact that $a=3$ alone tells you that the interval of integration starts at 3, and since we know the interval has length 5, that leaves the first choice as the correct answer.

6 0

3 years ago

Find 2 numbers a and b such that (a-b)^2 < (a+b)(a-b)< (a+b)^2 Find two numbes a and b such that this is not true

Svet_ta [14]

Suppose for the moment that the inequality holds for all $a,b$ :

$(a-b)^2$

Expanding everything gives

$a^2-2ab+b^2$

$\implies-ab$

In particular, the inequality says that $-ab$ for any choice of $a,b$ . But if $a$ and $b>0$ , then $-ab>0$ while $ab$ .

So one possible choice of $a,b$ could be $a=-1$ and $b=1$ . Then we get

$(-1-1)^2=4$

$(-1+1)(-1-1)=2$

$(-1+1)^2=0$

but clearly it's not true that $4$ .

8 0

3 years ago

The height of an object tossed upward with an initial velocity of 120 feet per second is given by the formula h = −16t2 + 120t,

AleksAgata [21]

Answer:

7.5 seconds

Step-by-step explanation:

$h = - 16 {t}^{2} + 120t \\ \because \: we \: have \: to \: find \: the \: time \: required \: \\ for \: the \: object \: to \: return to \: its \: \\ point \: of departure. \\ \therefore \: plug \: h = 0 \: in \: the \: given \: euation. \\ \therefore \: 0 = - 16 {t}^{2} + 120t \\ \therefore \:16 {t}^{2} - 120t = 0 \\ \therefore \:8t(2t - 15) = 0 \\ \therefore \:8t = 0 \: \: or \: \: (2t - 15) = 0 \\ \therefore \:t = \frac{0}{8} \: or \: t = \frac{15}{2} \\ \therefore \:t = 0 \: or \: t = 7.5 \\ \because \: t = 0 \: is \: not \: possible \\ \huge \purple{ \boxed{ \therefore \: t = 7.5 \: seconds}}$