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3 years ago

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A fire is sighted Due West of Lookout A. The bearing of the fire from Lookout B, 6.5 miles due south of A, is North 35°34'W. How

far is the fire from B (to the nearest tenth of a mile)?

1 answer:

KonstantinChe [14]3 years ago

4 0

Pm me i ne more info,,, is there some missing ?

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Need Help Please (b) Solve for x. Show each step of the solution. 3.5(7 – x) + 28 = 97 – 4.5(3x + 21)

leonid [27]

it is 13 1/3 . hope it helps

4 0

3 years ago

The distance between two points is 6_/6.One endpoint is (4,-1). It is known that the y-coordinate of the other endpoint is 9.

babunello [35]

The x-coordinate of the given endpoint will be " $x = 4 \pm 2\sqrt{29}$ ".

According to the question,

The end points are:

$(x_1,y_1) = (4,-1)$
$(x_2, y_2) = (x,9)$

Distance,

$6\sqrt{6}$

As we know,

The distance formula,

→ $Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

By substituting the values, we get

→ $6\sqrt{6} = \sqrt{(x-4)^2+(9+1)^2}$

→ $216= (x-4)^2+100$

→ $(x-4)^2 = 116$

→ $x-4 = \pm \sqrt{116}$

→ $x-4=\pm 2\sqrt{29}$

→ $x = 4 \pm 2\sqrt{29}$

Thus the above answer is right.

Learn more:

brainly.com/question/18310956

5 0

2 years ago

Plss help me plsss!!

grin007 [14]

Answer:

20 units

Step-by-step explanation:

The first part is already solved.

<u>We got the scale factor </u>

k = 4

<u>Use this to find the missing side:</u>

MN = k*YZ
MN = 4*5 = 20

3 0

3 years ago

Read 2 more answers

There are 3 different highways from City W to

devlian [24]

Answer:

6 x 4 you have 4 numbers and each can be put in 6 different ways. so it is 24 i think

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$

6 0

3 years ago