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3 years ago

10

Use graphs to determine which equation could possibly be an identity.

1 answer:

Rashid [163]3 years ago

3 0

Answer:

a. (sec(t)+csc(t))/(1+cot(t)) = sec(t)

Step-by-step explanation:

Subtracting the right side of the equation will result in an expression that is zero for all values of t if the original equation is an identity. That is the case for the equation of (a), but not for the others. See attached.

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What is the value of R

frutty [35]

Answer:

2.45 × $10^{6}$

Step-by-step explanation:

using the rules of exponents

• $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$

• $(a^m)^{n}$ = $a^{mn}$

evaluating the numerator

(3.8 × $10^{5}$ )² = 3.8² × $(10^5)^{2}$ = 14.44 × $10^{10}$

R = $\frac{14.44.10^{10} }{5.9.10^{4} }$

= $\frac{14.44}{5.9}$ × $\frac{10^{10} }{10^{4} }$

= 2.45 × $10^{6}$ ← to 2 dec. places

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The temperature started at 87*F . If the temperature wen down 10*F and up 17*F, whats is the temperature?

Alecsey [184]

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3 years ago

Help me with this please!!

valina [46]

<h2><u>Given:</u><u>-</u></h2>

Points C = (-7,2) → $\sf{(X_1,Y_1)}$
D = (3,12) → $\sf{(X_2,Y_2)}$

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

$\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}$

$→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}$

$→\;{\sf{\frac{-4}{2},\frac{14}{2}}}$

$→\;{\sf{-2,7}}$

The Midpoint of CD ◕➜ $\Large{\red{\mathfrak{(-2,7)}}}$

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → $\sf{(X_1,Y_1)}$

O = (-2,7) → $\sf{(X_2,Y_2)}$

$\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}$

$→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}$

$→\;{\sf{\sqrt{5²+5²}}}$

$→\;{\sf{\sqrt{25+25}}}$

$→\;{\sf{\sqrt{50}}}$

$→\;{\sf{5√2 (or) 7.07}}$

Radius of Circle ◕➜ $\Large{\red{\mathfrak{7.07}}}$

<h2>Option D.</h2>

Hope It Helps You ✌️

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lbvjy [14]

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