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3 years ago

Write an equation that involves multiplication, contains a variable and has a solution of 5

Mathematics

1 answer:

dolphi86 [110]3 years ago

6 0

B×3=5 3÷b=5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope that helped =)

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How did the war of 1812 start PLEAEE HELP NOW

Tanya [424]

Answer:

In 1812, with President Madison in office, Congress declared war against the British. The war began with an attack on Canada, both as an effort to gain land and to cut off British supply lines to Tecumseh's Indian confederation, which had long troubled the US. ... The US troops were badly routed.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm

jeyben [28]

Answer:

The length of the legs is 8.64cm and 14.64cm respectively

Step-by-step explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

$AB = 17$

$PAMC = 32$

$PBMC = 25$

From the attachment, we have:

$y + z = AB$

Since, M is the Midpoint

$y = z = \½AB$

Substitute 17 for AB

$y = z = \½ * 17$

$y = z = 8.5$

Also, from the attachment

$v + x + z = PAMC$

$v + x + y = 32$

Substitute 8.5 for y

$v + x + 8.5 = 32$

$v + x = 32 - 8.5$

$v + x = 23.5$ --------- (1)

Also, from the attachment

$v + w + z = 25$

Substitute 8.5 for z

$v + w + 8.5 = 25$

$v + w = 25 - 8.5$

$v + w = 17.5$ ----------- (2)

Subtract (2) from (1)

$v - v + x - w = 23.5 - 17.5$

$x - w = 6$

Make x the subject

$x = 6 + w$

Apply Pythagoras Theorem:

We have that:

$AB^2 = AC^2 + BC^2$

The above can be replaced with

$17^2 = x^2 + w^2$ (see attachment)

$289 = x^2 + w^2$

Substitute 6 + w for x

$289 = (6 + w)^2 + w^2$

$289 = 36 + 12w + w^2 + w^2$

$289 - 36 = 12w + 2w^2$

$253 = 12w + 2w^2$

Reorder

$2w^2 + 12w - 253 = 0$

Solve using quadratic equation:

$w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 2$

$b = 12$

$c = -253$

$w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}$

$w = \frac{-12 \± \sqrt{144 + 2024}}{4}$

$w = \frac{-12 \± \sqrt{2168}}{4}$

$w = \frac{-12 \± 46.56}{4}$

Split:

$w = \frac{-12 + 46.56}{4}$ or $w = \frac{-12 - 46.56}{4}$

$w = \frac{34.56}{4}$ or $w = \frac{-58.56}{4}$

$w = 8.64$ or $w = -14.64$

But length can't be negative

So:

$w = 8.64$

Recall that: $x = 6 + w$

$x = 6 + 8.64$

$x = 14.64$

Hence, the length of the legs is 8.64cm and 14.64cm respectively

5 0

3 years ago

On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters

inysia [295]

Answer: c. $P(I \text{ or }\overline{H})$

Step-by-step explanation:

Notation to show "not an event E" : $\overline{E}$

Given: I = the event that a player in an infielder.

H = the event that a player is a great hitter.

= the event that a player is not a great hitter.

The correct symbol for the probability that a player is an infielder or is not a great hitter : $P(I \text{ or }\overline{H})$

hence, the correct option is c. $P(I \text{ or }\overline{H})$

7 0

3 years ago

Find the number if: 4.5% of it is 23% of 45

Usimov [2.4K]

Well, let's say the number is "x", so "x" is the 100%.

now, what the dickens is 23% of 45 anyway?

if we take 45 to be the 100%, what is its 23%?

$\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
45&100\\
a&23
\end{array}\implies \cfrac{45}{a}=\cfrac{100}{23}\implies \cfrac{45\cdot 23}{100}=a\implies 10.35=a$

ok, well, we know then that 4.5% of "x" is 10.35, we also know that "x" is the 100%, what is "x" anyway?

$\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
x&100\\
10.35&4.5
\end{array}\implies \cfrac{x}{10.35}=\cfrac{100}{4.5}\implies x=\cfrac{10.35\cdot 100}{4.5}$

3 0

4 years ago

Read 2 more answers

Use the quadratic formula to find the solution to the question 3x^2-10x+5=0

nlexa [21]

Answer:

$x_{1} =$ $\frac{5 + \sqrt{10} }{3}$ and $x_{2} =$ $\frac{5 - \sqrt{10} }{3}$