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2 years ago

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many The scale of a map is 1 in. : 6 mi. How many inches does each distance occupy on the map? Round to the nearest tenth. 24. 2

3 mi 25. 10 mi 26. 45 mi 27. 16 mi

1 answer:

Amiraneli [1.4K]2 years ago

6 0

Answer:

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Step-by-step explanation:

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i need to know this ASAP i dont know how to do this can someone type in the answer step by step please

alexira [117]

M1+m2+m3=180
m5+m6+m7=180
Triangle angle sum theorem
m1+m2+m3=m5+m6+m7
Since m3 is congruent to m5
You can subtract m3 from the left and m5 from the right
m1+m2=m6+m7

3 0

2 years ago

Pran ran 29 miles less than Robin last week. Pran ran 19 miles. How many<br> miles did Robin run?

9966 [12]

Pran= 19
Robin= 29+19

29+19= 48

5 0

2 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant III and sec θ=61/60. Find the exact values of

Nataly_w [17]

Answer:

Part 1) $cos(\theta)=-\frac{60}{61}$

Part 2) $tan(\theta)=\frac{11}{60}$

Part 3) $cot(\theta)=\frac{60}{11}$

Part 4) $csc(\theta)=-\frac{61}{11}$

Part 5) $sin(\theta)=-\frac{11}{61}$

Step-by-step explanation:

we know that

If angle theta lie on Quadrant III

then

The function sine is negative

The function cosine is negative

The function tangent is positive

The function cotangent is positive

The function cosecant is negative

The function secant is negative

step 1

Find $cos(\theta)$

we know that

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

we have

$sec(\theta)=-\frac{61}{60}$ ----> the value must be negative

therefore

$cos(\theta)=-\frac{60}{61}$

step 2

Find $tan(\theta)$

we know that

$tan^{2} (\theta)+1=sec^{2} (\theta)$

we have

$sec(\theta)=-\frac{61}{60}$

substitute

$tan^{2} (\theta)+1=(-\frac{61}{60})^{2}$

$tan^{2} (\theta)+1=\frac{3,721}{3,600}$

$tan^{2} (\theta)=\frac{3,721}{3,600}-1$

$tan^{2} (\theta)=\frac{121}{3,600}$

$tan(\theta)=\frac{11}{60}$

step 3

Find $cot(\theta)$

we know that

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

we have

$tan(\theta)=\frac{11}{60}$

therefore

$cot(\theta)=\frac{60}{11}$

step 4

Find $csc(\theta)$

we know that

$cot^{2} (\theta)+1=csc^{2} (\theta)$

we have

$cot(\theta)=\frac{60}{11}$

substitute

$(\frac{60}{11})^{2}+1=csc^{2} (\theta)$

$\frac{3,600}{121}+1=csc^{2} (\theta)$

$\frac{3,721}{121}=csc^{2} (\theta)$

square root both sides

$csc(\theta)=-\frac{61}{11}$

step 5

Find $sin(\theta)$

we know that

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

we have

$csc(\theta)=-\frac{61}{11}$

therefore

$sin(\theta)=-\frac{11}{61}$

7 0

2 years ago

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Help pls ASAP I need it now plss

kumpel [21]

She or he payed
C. 18.80

8 0

2 years ago

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If f(x)=x³+x+1 and g(x)=-x, find (f°g)(x) and (g°f)(x).<br> (f°g)(x)=

andreev551 [17]

The value of (f· g)(x) from the given functions is $-x^4-x^2-x$ .

The given functions are f(x)=x³+x+1 and g(x)=-x.

<h3>What is the function?</h3>

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

Here,

(f·g)(x)= f(x) × g(x)

= (x³+x+1) × (-x)

= $-x^4-x^2-x$

(g·f)(x)= g(x) × f(x)

= (-x) × (x³+x+1)

= $-x^4-x^2-x$

Therefore, the value of (f· g)(x) from the given functions is $-x^4-x^2-x$ .

To learn more about the function visit:

brainly.com/question/28303908.

#SPJ1

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6 months ago