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geniusboy [140]
3 years ago
12

When q = 15, r = 4 Find q when r = 7

Mathematics
2 answers:
zhenek [66]3 years ago
7 0
The answer is18。。。。。
djyliett [7]3 years ago
6 0

Answer:

q = 18 ?

Step-by-step explanation:

This is not a complete question but ill give it a shot!

If we are following a pattern then we see q is 11 more than r.

So we do 11 + r (7) to get 18!

mark as brainliest

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The Controlled Substances Act of 1970 categorized drugs such as morphine and cocaine as Schedule ________ drugs. A. IV B. I C. I
Feliz [49]
I believe the answer is

C. II
8 0
2 years ago
3х – 6 =-2y<br> Find the slope of a line perpendicular to each given line
kondaur [170]

Answer:

Slope of the perpendicular line: 2/3

Step-by-step explanation:

-2y = 3x - 6

y = - 3/2x - - 6/2

y = -3/2 + 3

3 0
2 years ago
two ships leave a port at the same time. the first ship sails on a bearing of 55° at 12 knots (natural miles per hour) and the s
Korvikt [17]

Answer:

37.59 nautical miles

Explanation:

Distance = Speed x Time

The speed of the first ship = 12 knots

Thus, the distance covered after 1.5 hours

\begin{gathered} =12\times1.5 \\ =18\text{ miles} \end{gathered}

The speed of the second ship = 22 knots

Thus, the distance covered after 1.5 hours

\begin{gathered} =22\times1.5 \\ =33\text{ miles} \end{gathered}

The diagram representing the ship's path is drawn and attached below:

The angle at port = 90 degrees.

The triangle is a right triangle.

Using Pythagorean Theorem:

\begin{gathered} c^2=a^2+b^2 \\ c^2=18^2+33^2 \\ c^2=324+1089 \\ c^2=1413 \\ c=\sqrt[]{1413} \\ c=37.59\text{ miles} \end{gathered}

The two ships are 37.59 nautical miles apart after 1.5 hours.

8 0
1 year ago
In ⊙H, Arc I K ≅ Arc J K, mArc I K = (11x + 2)°, and mArc J K = (12x – 7)°.
Bumek [7]

Answer:

202

Step-by-step explanation:

Did it on Edu.

7 0
3 years ago
Read 2 more answers
Can someone give me the answers asap! 15 points
Eduardwww [97]
In the first question,
Since D is the midpoint of line segment AC.
therefore, AD=DC.

In triangles, ABD & CBD.
Since, AB is congruent to CB and BD is a common side between those two triangles.
Therefore, AD is congruent to DC.

Finally, in triangles DEA & DEC
Since, ED line segment is a common side between those triangles and the measure of angle EDA is as the same of angle EDC which are equal to 90 degrees and AD is congruent to DC.
Therefore, triangle EDA is congruent to triangle EDC ( you have to take care of the triangle order letters in congruence) and EC is congruent to EA.
3 0
2 years ago
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