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3 years ago

I need help on the Math question?

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

3 0

Answer:

The answer is D.

Step-by-step explanation:

3x*3x = 9x^2

3x*2= 6x

5*3x=15x

5*2=10

6x+15x=21x

9x^2+21x+10

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So I’m not smart so like help meh pleasee<br><br> What is 49% of 200?

____ [38]

Answer:98

Step-by-step explanation:

We assume, that the number 200 is 100%.if 200 is 100%, so we can write it down as 200=100%.We also know, that x is 49% of the output value, so we can write it down as x=49%.Now we have two simple equations:

1) 200=100%

2) x=49%

we multiply both sides of the equation by x.then we divide both sides of the equation.

3 0

3 years ago

Your cousin borrowed $6000, some on a home-equity loan at an interest rate of 9.5% and the rest on a consumer loan at an interes

Novosadov [1.4K]

Answer:

I=PRT 9.5% in decimal form is .095=11 times rate times tax but we don't know what the tax is.

6 0

3 years ago

If DM = 35, what is the value of r?

Gennadij [26K]

Answer:

$r = 11$

Step-by-step explanation:

Given

$DM = 35$

$DG = r + 5$

$GM = 3r - 14$

Required

Find r

From the question, we understand that G is a point between D and M:

This implies that:

$DM = DG + GM$

Substitute values for DM, DG and GM

$35 = r + 5 + 3r - 14$

Collect Like Terms

$r + 3r = 35 - 5 + 14$

$4r = 44$

Solve for r

$r = 44/4$

$r = 11$

3 0

3 years ago

25 points please help me!

lianna [129]

$1.14 1.40 x 0.8=1.12 1.4+ 1.12=1.52 * 0.25 = 1.14

8 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: