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lawyer [7]
2 years ago
9

Bill rode 9 miles on saturday. 9n sunday he rode his bicycle 4 times the number of miles he rode on saturday.

Mathematics
2 answers:
pentagon [3]2 years ago
4 0
The answer to the problem is 36.

elena-s [515]2 years ago
3 0
He rode 36 miles on sunday
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If tsr = 84 what is the value of x? SQ bisects TSR RSQ= 3x-9
dsp73
We know that
 ∠ TSR = 84°

if SQ bisects ∠ <span>TSR
then 
</span>∠ RSQ = ∠ TSR/2
<span>so
</span>∠ RSQ = (1/2)*84°----- 42°
∠ RSQ = 3x-9
3x-9=42-------> 3x=42+9------> 3x=51-----> x=51/3-----> x=17°
<span>
the answer is
</span>x=17°<span>

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7 0
3 years ago
Read 2 more answers
Paul has 16 fish pictures and 12 deer pictures. He wants to arrange the pictures in rows with the same numbers of fish and of de
VladimirAG [237]

Answer:

4 rows

Step-by-step explanation:

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3 years ago
Que costaría más comprar una vaca o comprar hamburguesas de 1€ hasta igualar a una vaca.
butalik [34]

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Step-by-step explanation:

8 0
2 years ago
Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!
katrin2010 [14]

Answer:

  13.98 in²

Step-by-step explanation:

I don't understand it, either.

Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.

__

The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.

The area of a segment is given by the formula ...

  A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.

Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...

  A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²

Rounded to hundredths, this is ...

  ≈ 13.98 in²

3 0
3 years ago
Please I need help on solving for x
Sedaia [141]

Answer:

34^o=1/2((3x-2)^0-(x+6)^o)

68^o=3x^o-2^o-x^o-6^o

68^o+8^o=2x^o-8+8^o

76^o=2x^o-->x=38^o

\huge\boxed{\boxed{\underline{\textsf{\textbf{x=38}}}}}

<h3><u>---------------------</u></h3><h3><u>hope it helps...</u></h3><h3><u>have a great day!!</u></h3>

5 0
2 years ago
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