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3 years ago

The diagonals of a square are___.

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

7 0

Answer:

The diagonals of a square intersect cross in a 90 degree angle. This means that the diagonals of a square are perpendicular. The diagonals of a square are the same length congruent.

Step-by-step explanation:

so the answer is perpendicular

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In the triangle below, what is the length of BC ? need help?!

Artemon [7]

Answer:

Step-by-step explanation:

just look at the other side since both are the same

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3 years ago

Need help with this question plz

neonofarm [45]

Answer:

A (3,-2)

Step-by-step explanation:

x = t-1

y = -sqrt(t)

Let t= 4

x = 4-1 = 3

t = -sqrt(4) = -2

The (x,y) coordinate is (3,-2)

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3 years ago

WILL GIVE BRAINLIST .. . .

Jobisdone [24]

Yep it’s B or 20 outcomes

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3 years ago

The number of bacteria in a refrigerated food product is given by N ( T ) = 22 T 2 − 123 T + 40 , 6 < T < 36 , where T is

antoniya [11.8K]

Answer:

$N(T(t)) = 1408t^2 - 385.6t - 105.52$

Time for bacteria count reaching 8019: t = 2.543 hours

Step-by-step explanation:

To find the composite function N(T(t)), we just need to use the value of T(t) for each T in the function N(T). So we have that:

$N(T(t)) = 22 * (8t + 1.7)^2 - 123 * (8t + 1.7) + 40$

$N(T(t)) = 22 * (64t^2 + 27.2t + 2.89) - 984t - 209.1 + 40$

$N(T(t)) = 1408t^2 + 598.4t + 63.58 - 984t - 169.1$

$N(T(t)) = 1408t^2 - 385.6t - 105.52$

Now, to find the time when the bacteria count reaches 8019, we just need to use N(T(t)) = 8019 and then find the value of t:

$8019 = 1408t^2 - 385.6t - 105.52$

$1408t^2 - 385.6t - 8124.52 = 0$

Solving this quadratic equation, we have that t = 2.543 hours, so that is the time needed to the bacteria count reaching 8019.

4 0

4 years ago

Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

Art [367]

Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

4 0

3 years ago