[tex]\text{L.H.S}\\\\=4\sin(-x) \cos(-x)\\\\=-4\sin x \cos x\\\\=-2 \cdot 2 \sin x \cos x\\\\=-2 \sin 2x\\\\=\text{R.H.S}\\\\\text{The identity is true.}[/tex]
A 20-ounce bottle of lotion costs $3.95. How much does each ounce cost?\
Answer:
$0.20
Step-by-step explanation:
3.95/20 = 0.1975
0.1975 would round up to 0.20
Answer:
$0.1975
Step-by-step explanation:
Do $3.95 divided by 20 so you can find the proportion of 1 ounce
How do you check the apparent solution of the following systems of equations: x+3y=8, x-4y=-6
Answer:
x=2
y=2
Step-by-step explanation:
x+3y=8
- x-4y=-6
7y=14
y=14/7
y=2
x+3(2)=8
x+6=8
x=8-6
x=2
y=2
x=2
in the lab, Kareem has two solutions that contain alcohol and is mixing them with each other. Solution A is 2% alcohol and Solution B is 7% alcohol. He uses 1200 milliliters of Solution A. How many milliliters of Solution B does he use, if the resulting mixture is a 4% alcohol solution?
Answer:
800ml
Step-by-step explanation:
solution A= 2% alcohol
solution B=7% alcohol
1200ml=1.2L
1200ml=1200cm^3
X= solution B
solution A + solution B=(1200+x)cm^3
% alcohol= amount of alcohol/total solution(AorB)×100
☆in 1200ml of solution A there's
0.02x1200
2x12
=24ml of alcohol (in 1200ml of solution A)
0.07x X
=0.07Xml of alcohol (in some ml of solution B)
(24+0.07X)ml of alcohol in = solution A and solution B
solution A + solution B= 1200+X
0.04 (1200+X)=24+0.07X
48+0.04X=24+0.07X
48-24=0.07X-0.04X
24=0.03X
X=800ml
What is the value of g?
Answer:
90
Step-by-step explanation:
I'm not sure about the answer
What is the volume of a sphere with a diameter of 50 cm
since we know the diameter is 50, then its radius is half that or 25.
[tex]\textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=25 \end{cases}\implies \begin{array}{llll} V=\cfrac{4\pi (25)^3}{3}\implies V=\cfrac{62500\pi }{3} \\\\\\ V\approx 65449.85~cm^3 \end{array}[/tex]
Our school’s girls volleyball team has 14 players, including a set of
3 triplets: Alicia, Amanda, and Anna. In how many ways can we
choose 6 starters if at most one of the triplets is in the starting lineup? There can't be 2 or more triplets and there can be none.
Answer:
[tex]1,\!848[/tex].
Step-by-step explanation:
There are two disjoint sets of ways to choose a lineup as required:
Include none of Alicia, Amanda, or Anna, orInclude exactly one of Alicia, Amanda, and Anna.Assume that none of Alicia, Amanda, or Anna is to be selected. This lineup of [tex]6[/tex] would then need to be selected from a set of [tex]14 - 3 = 11[/tex] players (which excludes Alicia, Amanda, and Anna.)
The number of ways of selecting (without order) [tex]6[/tex] items out of a set of [tex]11[/tex] (distinct) items is equal to the combination:
[tex]\begin{aligned}\begin{pmatrix}11 \\ 6\end{pmatrix} &= \frac{11!}{(6!)\, (11 - 6)!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}[/tex].
Assume that Alicia is selected, but neither Amanda nor Anna is selected. The other [tex]6 - 1 = 5[/tex] players in this lineup would then need to be selected from a set of [tex]14 - 1 - 2 = 11[/tex] players. (This set of [tex]11[/tex] excludes Alicia, Amanda, and Anna.)
The number of ways to select [tex]5[/tex] items from a set of [tex]11[/tex] items is:
[tex]\begin{aligned}\begin{pmatrix}11 \\ 5\end{pmatrix} &= \frac{11!}{(5!)\, (11 - 5)!} \\ &= \frac{11!}{5! \times 6!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}[/tex].
Similarly, there would be another set of [tex](11!) / (6! \times 5!)[/tex] distinct ways to select the lineup if Amanda is selected, but neither Alicia nor Anna is.
Likewise, the number of ways to select the lineup with Anna but neither Amanda nor Alicia would also be [tex](11!) / (6! \times 5!)[/tex].
These sets of configurations for the lineup are pairwise disjoint from one another. Thus, the total number of ways to select this lineup would be:
[tex]\begin{aligned}& \begin{pmatrix}11 \\ 6 \end{pmatrix} + 3 \times \begin{pmatrix}11 \\ 5 \end{pmatrix} \\ =\; & \frac{11!}{6! \times 5!} + 3 \times \frac{11!}{6! \times 5!} \\ =\; & \frac{4 \times 11!}{6! \times 5!} \\ =\; & \frac{4 \times 11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \\ =\; & \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 3 \times 2} \\ =\; & 1,\!848\end{aligned}[/tex].