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4 years ago

Someone PLS!! Help me. And do the one before it.

Mathematics

1 answer:

My name is Ann [436]4 years ago

6 0

The answer is b because one bottle is 2.50 and they bought 5 then it says they bought 2 scarfs for 2 for 10.

Final answer: B

Hope this helped :)

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I do not need an explanation, just a simple angle number!

Murljashka [212]

Answer:

Step-by-step explanation:

a + 143 = 180 {Linear pair}

a = 180 - 143

a = 37

37 + 93 + b = 180 {Angle sum property of triangle}

130 + b = 180

b = 180 - 130

b = 40

c + b = 90

c + 40 = 90

c = 90 - 40

c = 50

d + 120 = 180 {Linear pair}

d = 180 - 120

d = 60

? = 60 + 40 {Exterior angle property of triangle}

? = 100

3 0

2 years ago

The graph of the function f(x)=−4(x−3)2+9 will be a parabola opening in which direction?

sesenic [268]

Answer:

downwards

Step-by-step explanation:

the equation of a parabola in vertex form is

f(x) = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

• if a > 0 then parabola opens upwards

• if a < 0 then parabola opens downwards

for

f(x) = - 4(x - 3)² + 9 with a = - 4 < 0 , then

the parabola opens downwards

8 0

2 years ago

What is m+7=24 I will give brainlist!

Genrish500 [490]

Answer:

m= 234

Step-by-step explanation:

3 0

3 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities