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3 years ago

According to the diagram what is cos 23 degrees?

1 answer:

8 0

.9205 or so is what it should say. (I don't have the graph on me.) and to read it to find that, find the angle measurement (23 in this case) then go to the column that says cos on it and read the number. It's that simple:) Hope this helps:)

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Find the 50th term of this arithmetic sequence: 6, 13, 20, 27, . . .

sladkih [1.3K]

Answer: 349

Step-by-step explanation:

Using the arithmetic sequence, it would add up to be 349

7 0

2 years ago

Johanna rode her bike 7 miles to a lake. She rode 3/4 of the distance to the lake before lunch. How many miles did Johanna ride

Alex777 [14]

Answer:

5.25 miles

Step-by-step explanation:

7 0

4 years ago

Can you please help me i don't understand this

inysia [295]

Answer:

10) Supplementary; 11) Supplementary; 12) Complementary

Step-by-step explanation:

Sum of supplementary angles equals 180° while sum of complementary angles equals 90°.

For question 10

x + 70° = 180° (sum of straight lines)

x = 180° - 70° = 110°

Hence, 70° and 110° are supplementary angles.

For question 11,

x + 85° = 180° (sum of straight lines)

x = 180° - 85° = 95°

Hence, 85° and 95° are supplementary angles.

For question 12,

x + 32° = 90° (sum of right angles)

x = 90° - 32° = 58°

Hence, 32° and 58° are complementary angles.

4 0

3 years ago

Where r is the radius of the cylinder and h is the height of the cylinder.

AnnyKZ [126]

<h3>Given:-</h3>

radius = 7 inches
height = 9 inches

$\\ \\$

Surface area of cylinder

$\\ \\$

we know:-

$\bigstar \boxed{ \rm Surface \: Area \: of \: Cylinder=2\pi rh+2\pi {r}^{2} }$

$\\ \\$

So:-

⇢SA of cylinder=2πrh+2πr²

$\\$

⇢SA of cylinder = 2πr(h+r)

$\\$

⇢SA of cylinder = 2 × π × 7(7 + 9)

$\\$

⇢SA of cylinder = 2 × π × 7 (16)

$\\$

⇢SA of cylinder = 2 × π × 112

$\\$

⇢SA of cylinder = 2 × 112 × π

$\\$

⇢SA of cylinder = 224π inches ²

$\\$

.°. Surface Area of cylinder is 224π inches².

3 0

2 years ago

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

Svetllana [295]

Naturally, any integer $n$ larger than 127 will return $127\equiv127\mod n$ , and of course $127\equiv0\mod127$ , so we restrict the possible solutions to $1\le n$ .

Now,

$127\equiv7\mod n$

is the same as saying there exists some integer $k$ such that

$127=nk+7$

We have

$\implies 120=nk$

which means that any $n$ that satisfies the modular equivalence must be a divisor of 120, of which there are 16: $\{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}$ .

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

$127=21\cdot6+1\iff127\equiv1\equiv7\mod6$

(If we're allowing $n=1$ , then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)

5 0

4 years ago