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Romashka [77]
3 years ago
8

What does 264÷58 equal to​

Mathematics
2 answers:
aalyn [17]3 years ago
5 0

Answer:

4.55

Step-by-step explanation:

264÷58= 4.55

Rudik [331]3 years ago
4 0
4.55 because u just divide them together do all the work and thats what u get
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consider the point A= (1,2) and the line B, given by the equation y=3x-5. write an equation in slope intercept of the line passi
Gre4nikov [31]

Answer:

y = 3x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 3x - 5 ← is in slope- intercept form

with slope m = 3

Parallel lines have equal slopes, thus

y = 3x + c ← is the partial equation of the parallel line

To find c substitute (1, 2) into the partial equation

2 = 3 + c ⇒ c = 2 - 3 = - 1

y = 3x - 1 ← equation of parallel line

5 0
3 years ago
A girl is 6 years younger than her brother.the product of their ages is 135.find their ages<br>​
Vaselesa [24]

Answer:

Brother's age = 15

Sister's age = 9

Step-by-step explanation:

4 0
2 years ago
Evaluate the expression for c = 6/7, 54/21 divided by c
Degger [83]
This is how i worked it out.
Good Luck

4 0
3 years ago
Read 2 more answers
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
There are some counters in a bag.
iVinArrow [24]

Answer:

187

Step-by-step explanation:

4 0
2 years ago
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