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3 years ago

Help please and thank you

Mathematics

2 answers:

Lana71 [14]3 years ago

5 0

Answer:

q=6

Step-by-step explanation:

Alexeev081 [22]3 years ago

5 0

Solution is here:
so q=-19

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If a+b=5 and a-b=4 then b^2-a^2 =

Anna007 [38]

Answer:

$a + b = 5 \\ a \: = 5 - b \\ a - b = 4 \\ 5 - b - b = 4 \\ 5 - 2b = 4 \\ 5 - 4 = 2b \\ 1 = 2b \\ \frac{1}{2} = \frac{2b}{2} \\0.5 = b \\ b = 0.5 \\ a = 5 - b \\ = 5 - 0.5 \\ = 4.5 \\ {a \: }^{2} - {b}^{2} = ( {4.5}^{2}) - ( {0.5}^{2} ) \\ = 20.25 - 0.25 \\ = 20 \\ {a}^{2} - {b}^{2} = 20$

Step-by-step explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

5 0

3 years ago

A number that divided by 3 leaves 2, divided by 5 leaves 3 and divided by 7 leaves 2. What is the number?

Bogdan [553]

X/3=2
x=6

x/5=3
x=15

x/7=2
x=14

Each time, you multiply both sides by the denominator.

Hope this helps :)

7 0

3 years ago

What is 4/5×24/12 equal to and simplified

klasskru [66]

It should be 8/5 cause 4/5 times 24/12= 4/5 times 2/1. then u multiply.

4 0

3 years ago

Adam buys a car for £15060 which depreciates in value at a rate of 1.5 per year. Work out how much Adam's car will be worth in 9

Step2247 [10]

Answer: £13026.90

Step-by-step explanation:

1.5% of £15060 = 225.90

225.90 x 9 = 2033.10

15060 - 2033.10 = 13026.90

hope this helps ;)

4 0

3 years ago

Read 2 more answers

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago