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3 years ago

Part a : solve - vp + 40 < 65 for v .

Mathematics

1 answer:

miskamm [114]3 years ago

6 0

Answer:

$\large \text{Part a:}\\\\for\ d-\dfrac{25}{d}}$

$\large\text{Part b:}\\\\\boxed{r=\dfrac{7}{3}w-5}$

Step-by-step explanation:

$\text{Part a}\\\\-vp+400,\ \text{then}\ \boxed{v>-\dfrac{25}{d}}\\\\\text{if}\ d$

$\text{Part b}\\\\7w-3r=15\qquad\text{subtract 7w from both sides}\\\\-3r=-7w+15\qquad\text{divide both sides by (-3)}\\\\\boxed{r=\dfrac{7}{3}w-5}$

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The amount that results when $4,000 is compounded at 6% annually over seven years.

DerKrebs [107]

Answer:$ 6,014.52

Step-by-step explanation:

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3 years ago

How many words of five letters can be created that start with the letter T, contain 2 other consonats and end with 2 vowels? Not

Jet001 [13]

If we're only counting 5 vowels (A, E, I, O, U) and 20 consonants (everything else, minus T), then there are

$\dbinom52=\dfrac{5!}{2!(5-2)!}=10$

ways of picking the vowels, and

$\dbinom{20}2=\dfrac{20!}{2!(20-2)!}=190$

ways of picking the consonants.

We want the word to start with T, and we'll allow any arrangement of the other 4 letters, so that the total number of words is

$4!\dbinom52\dbinom{20}2=\boxed{45,600}$

Keep in mind that this means words like TRIES and TIRES are treated as different.

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3 years ago

Which property of multiplication is shown by the equation f*8=8f

Ivanshal [37]

Commutative Property of Multiplication.

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3 years ago

I need to know the improper fractions answers for:

zmey [24]

Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

Part 4) $b= \frac{45}{4}\ units$

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

substitute

$\frac{1}{2}=\frac{x}{15}$

solve for x

$x=\frac{15}{2}\ units$ ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

$z^2=225-\frac{225}{4}$

$z^2=\frac{675}{4}$

$z=\frac{\sqrt{675}}{2}\ units$

simplify

$z=\frac{15\sqrt{3}}{2}\ units$

step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}$

$b= \frac{45}{4}\ units$

7 0

3 years ago

Determine whether you would use the Law of Sines or the Law of Cosines to solve each triangle. Do not solve the triangle.

Andrej [43]

A triangle is a three-edged polygon with three vertices. Only the law of Sine can be applied to solve the triangle.

<h3>What is a triangle?</h3>

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°.

The sum of all three angles can be written as,

118° + 22° + x = 180°

x = 40°

For the sine rule to be applicable, two angles and a side or an angle and two sides are needed. Therefore, the law of Sine can be applied to solve the triangle.

For the cosine rule to be applicable, two sides and an angle between them is needed to be known, hence, the law of cosine can not be applied.

Hence, Only the law of Sine can be applied to solve the triangle.

Learn more about Triangle:

brainly.com/question/2773823

#SPJ1

6 0

2 years ago