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3 years ago

How should the subject-verb agreement in this sentence be changed, if at all?

Mathematics

1 answer:

nikdorinn [45]3 years ago

4 0

The answer will be neither the bus nor the trains were running today. (D)

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Can someone help me

nikdorinn [45]

So you will put the time and distance together like this (4,3) (8,6) (12,9) (16,12) (20,5)

6 0

3 years ago

Will mark brainliest. <br> Find cos(a-(π/4)), if cos(a)= -(1/3)

grigory [225]

Answer:

Step-by-step explanation:

$cos(\frac{-1}{3}-\frac{\pi }{4})$

$cos(\frac{-4}{12}-\frac{3\pi }{12})$

$cos(\frac{-13.42}{12})$ = 0.44 <- rounded to the nearest hundredth

6 0

4 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

A party mix is made by adding nuts that sell for $2.50 per kg to a cereal mixture that sells for $1 per kg. How much of each sho

Nana76 [90]

Answer:

n=16 kg

c=44 kg

Step-by-step explanation:

Let

n=nuts at $2.50 per kg

c=cereal at $1 per kg

n+c=60 (1)

2.5n + c=1.40(60) (2)

From (1)

n=60-c

Substitute n=60-c into (2)

2.5n + c=1.40(60)

2.5(60-c) +c=84

150-2.5c+c=84

-1.5c=84-150

-1.5c= -66

Divide both sides by -1.5

c=44 kg

Substitute c=44 into (1)

n+c=60

n+44=60

n=60-44

=16

n=16 kg

The amount of nuts is 16 kg and cereal mixture is 44 kg to obtain 60 kg of a mix that will sell for $1.40 per kg.

8 0

3 years ago

The degree of a vertex is <br><br> Help please

Nina [5.8K]

Answer:

is the number of graph edges which touch v

Step-by-step explanation:

To find the degree of a graph, figure out all of the vertex degrees. The degree of the graph will be its largest vertex degree. The degree of the network is 5.

Once you know the degree of the verticies we can tell if the graph is a traversable by lookin at odd and even vertecies. First lets look how you tell if a vertex is even or odd.

7 0

3 years ago