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3 years ago

3 plus the product of 2 and h

Mathematics

1 answer:

Ket [755]3 years ago

4 0

I'm pretty sure it should be 2h+3

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What are the partial products of 42x5?

polet [3.4K]

(40×5)+(2×5) so 40×5=200 and 2×5=10 so 200+10=210

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2 years ago

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Select the interval where f(x)>0<br><br> A: -2 B: 1< x<2<br> C: 4<br> HELP

Viefleur [7K]

Answer: 1 < x < 2

Step-by-step explanation:

7 0

3 years ago

Find the area of parallelogram ABCD given m A = 30 and the following measures. AB = 6 ft.; AX = 3√3 A = ? A. 18 ft.² B. 18√3 ft.

S_A_V [24]

Answer:

Given that a parallelogram ABCD has sides AB= 6 ft and AX = 3√3

Angle A =30

To find the area of the parallelogram

We have area of parallelogram = base x height

Here base can be taken as side AB = 6 ft

Height = perpendicular distance of AB from vertex X

= AX sin A

= 3√3sin30

=1.5√3

Hence area =6(1.5√3)=9√3ft^2

Read more on Brainly.com - brainly.com/question/12140514#readmore

Step-by-step explanation:

6 0

2 years ago

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Write <, >, or = to make the statement true : .293 ___ .29

JulsSmile [24]

If you add a zero to the end of .29 it becomes .290
So, the answer is .293 > .29

7 0

3 years ago

A petrol kiosk p is 12 km due north of another petrol kiosk q. The bearing of a police station r from p is 135 degree and that f

tiny-mole [99]

Answer:

Distance between P and R is 40.15 km.

Step-by-step explanation:

From the picture attached,

Petrol kiosk P is 12 km due North of another petrol kiosk Q.

Bearing of a police station R is 135° from P and 120° from Q.

m∠QPR = 180° - 135° = 45°

m∠PQR = 120°

m∠PRQ = 180° - (m∠QPR +m∠PQR)

= 180° - (45° + 120°)

= 180° - 165°

= 15°

Now we apply sine rule in ΔPQR to measure the distance between P and R.

$\frac{\text{sin}(\angle QPR)}{\text{QR}}= \frac{\text{sin}(\angle PQR)}{\text{PR}}=\frac{\text{sin}\angle PRQ}{\text{PQ}}$

$\frac{\text{sin}(45)}{\text{QR}}= \frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}$

$\frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}$

PR = $\frac{12\text{sin}(120)}{\text{sin}(15)}$

PR = 40.15 km

Therefore, distance between P and R is 40.15 km.

8 0

3 years ago