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3 years ago

Write the numerical expression for “subtract 3 from 5”.

Mathematics

2 answers:

KatRina [158]3 years ago

7 0

Answer:

5 - 3

Step-by-step explanation:

marishachu [46]3 years ago

6 0

To subtract 3 from 5, you would write 5 - 3

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What is the value of the expression -5(3+4)?

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-35 i think it is tha

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Kryger [21]

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3 years ago

How much does a 24 case of 16.9 oz bottles water weigh?

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3 years ago

a square is put together by two triangles. the hypotenuse is 11 cm and width if 8 cm. find the height of the triangle.

maksim [4K]

Answer:

b ≈ 7.54 cm

Step-by-step explanation:

Pythagorean Theorem

a² + b² = c²

a and b are the sides and c is the hypotenuse

hypotenuse is 11 cm

a side is 8 cm

8² + b² = 11²

64 + b² = 121

-64 -64

b² = 57

√b² = √57

b ≈ 7.54 cm ( use ≈ because its rounded so its not exact)

3 0

4 years ago

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago