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2 years ago

Helppppppp Plzzzzzzzzzzz

Mathematics

1 answer:

Fudgin [204]2 years ago

5 0

Step-by-step explanation:

ill help you in a little gimmie a sec

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The LCM of 2^2×5^2×7 and another number N is 2^2×3×5^2×7^2 find the lowest value of N

Naddika [18.5K]

256: 2(8) :)))))))))))))))))

4 0

2 years ago

Read 2 more answers

Please urgent fast thanks

stich3 [128]

Answer:

A) I = $171.73

B) $4771.73

Step-by-step explanation:

We are told her loan is $4600 for 7 months.

Thus;

Principal; P = 4600

Time; T = 7 months = 7/12 year

Interest rate is given as 6.4%

Thus; r = 6.4

A) To get the interest, the Formula is;

I = PRT/100

I = (4600 × 6.4 × (7/12))/100

I = $171.73

B) Since interest is $171.73, it means amount she owes after the 7 months = $4600 + $171.73 = $4771.73

6 0

2 years ago

Please help!!<br> Each pair of polygons is similar. Find the value of x.

Simora [160]

Answer in fraction form: x = 55/4

Answer in decimal form: x = 13.75

Both forms are equivalent, but they say it in a slightly different way.

===========================================================

Work Shown:

AC/QY = CD/YW

(x+5)/15 = 5/4

4(x+5) = 15*5

4x+20 = 75

4x = 75-20

4x = 55

x = 55/4

x = 13.75

------------

Explanations:

The first step is possible because the polygons are similar. Similar figures have their corresponding sides be in the same proportion or ratio. Note how side AC has the double red angle (at point A) and a single red angle (point C). The side QY has the same idea in the same order. This is why AC corresponds to QY. The ratio AC/QY is the same as CD/YW.
In the second step, I plugged in the expressions that the diagram shows. This is the substitution property.
On step 3, I cross multiplied. The general format is A/B = C/D cross multiplies to A*D = B*C.

7 0

3 years ago

Read 2 more answers

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$