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2 years ago

A rhombus has its diagonals 7cm and 6 cm whats its area

Mathematics

1 answer:

ipn [44]2 years ago

5 0

$\large\frak{ Given :}$

Diagonal of a rhombus = 7 cm and 6 cm

$\large\frak{Find:}$

Area of a rhombus

$\large\frak{Explanation :}$

$\small\tt Area = \frac{1}{2}× d¹ × d²$

Where,

d¹ = length of diagonal 1
d² = length of diagonal 2

$\implies \small\tt{ \frac{1}{2} \times 7 \times 6}$

$\implies\small\tt{ \frac{1}{2} \times 42 }$

$\implies \small\tt{Area = 21 {cm}^{2} }$

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Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?

Alex Ar [27]

Answer:

$\frac{x^2}{12}-\frac{y^2}{4}=1$

Step-by-step explanation:

Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:

Formula for hyperbola with horizontal transverse axis centered at origin

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Directrices -> $x=\pm\frac{a^2}{c}$
Foci -> $(\pm c,0)$ where $a^2+b^2=c^2$
$a>b$

Since we are given our directrices of $x=\pm3$ and foci of $(\pm4,0)$ , then we can set up the directrices equation to solve for $a^2$ :

$x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2$

Now we can determine $b^2$ to complete our equation for the hyperbola:

$a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4$

Therefore, our equation for our hyperbola is $\frac{x^2}{12}-\frac{y^2}{4}=1$

5 0

2 years ago

#14 i

ivann1987 [24]

Answer:

Incorrect/No

Step-by-step explanation:

7x2=14

-7x2=-14

Follow the below guidelines:

A positive number times a positive number is a positive number
A negative number times a positive number is a negative number
A positive number times a negative number is a negative number
A negative number times a negative number is a positive number

Looking at the last one, we see that -2x-7=14, a positive number

Hope this helps!

--Applepi101

7 0

3 years ago

If the range of the function y = f(x) is {y | y ≥ 11, y ∈ R}, then the range of the new function g(x) = f(x + 2) - 3 is

mafiozo [28]

Answer:

y ≥ 8

Explanation:

Note that if f(x) is transformed into f(x + a) - b

The original functions is shifted a units to the left and b units downward.

Horizontal shifting will not affect the range of the function, only vertical shifting will change its range.

From the given, g(x) is the transformation of f(x) with

$g(x)=f(x+2)-3_{}$

f(x) is shifted 2 units to the left and 3 units downward, we will disregard the horizontal shifting.

Since f(x) has a range of y ≥ 11, and g(x) is 3 units downward, the range will also move 3 units downward.

y ≥ 11 - 3

The answer is y ≥ 8

3 0

1 year ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago

A farmer has 150 rabbits. There are 4 times as many female rabbits as male rabbits. How many male and female rabbits are there?

enyata [817]

There are 120 female rabbits and 30 male rabbits because 30 times 4 is 120 and 120 plus 30 is 150

7 0

3 years ago