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2 years ago

13

What would be the range of possible measures of 5 and 6

2 answers:

Scrat [10]2 years ago

8 0

Answer:

1

Step-by-step explanation:

vazorg [7]2 years ago

6 0

The answers would be 1

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How to solve -2(x+3)=8

Shtirlitz [24]

-2(x + 3) = 8

Multiply -2 with x and 3

-2x -6 = 8

Now add 6 to both sides

-2x -6 = 8
+6 +8

-2x = 16

Lastly divide -2 from both sides to find x

-2x = 16
—— —-
-2x -2x

X = -8

5 0

2 years ago

Read 2 more answers

Jenny spins a fair 4-sided spinner what is the probabiliity that the spinner will land on a or b

algol [13]

We require a picture of said spinner to see which sides say a and which say b

6 0

1 year ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

2. What is an equation line that pass-cs through the point (3,-1) and h (4)

joja [24]

Y=mx+b
m=slope
b=y intercept
we are given (3,-1)
one solution is x=3 and y=-1 or
when x=3 then y=-1
also slope=2
so
y=2x+b
subsitue 3 for x and -1 for y and solve for b
-1=2(3)+b
-1=6+b
subtract 6
-7=b

y=2x-7 is equation

answer is number 4

5 0

2 years ago

The diagram shows a prism work out the volume???

julia-pushkina [17]

Answer:

Volume is equal to

L×W×H

=(10×9×7)cm

=90×7

=630

8 0

3 years ago