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3 years ago

Can you use PEMDAS without parenthesis, exponents, division, or subtraction? Example: 5+1x10=?

2 answers:

7 0

If there are that many things missing then go left to right; 5+1 then multiply the sum by ten

5+1=6
6x10=60

I hope I was helpful

Kaylis [27]3 years ago

4 0

Yes just use wat you learned with PEMDAS
Since you dont have parenthesis or exponents or the division
just do 1x10 first and you get 10 then you do 5+10 = 15
so for the example you set it would be 15

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an experiment consists of flipping a fair coin 4 times. What is the probability of obtaining at least one head?

Dvinal [7]

$\mathbb P(X\ge1)=1-\mathbb P(X=0)$

The probability of getting 0 heads in 4 tosses (or equivalently, 4 tails) is $\dfrac1{2^4}=\dfrac1{16}$ .

So the desired probability is

$1-\dfrac1{16}=\dfrac{15}{16}$

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3 years ago

PLEASE HELP !!!!!!! For what value of x is line m parallel to line n? Enter your answer in the box. x = Line m is parallel to li

Sonja [21]

Answer:

$x=10$

Step-by-step explanation:

we know that

When two lines are crossed by another line (transversal), the angles in matching corners are called Corresponding Angles.

When the line are parallel the corresponding angles are equal in measurement.

In this problem

$(8x+50)\°=130\°$ -----> by corresponding angles

see the attached figure to better understand the problem

Solve for x

Subtract 50 both sides

$8x=130-50$

$8x=80$

Divide by 8 both sides

$x=80/8$

$x=10$

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3 years ago

Round 23,421 to the nearest thousand and ten thousand

Ronch [10]

To the nearest thousand is 23,000 and the nearest ten thousand is 20,000

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3 years ago

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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°

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3 years ago