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3 years ago

Write with explain.

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

8 0

Answer: The acute angle between diagonals is 50

Step-by-step explanation:

Red angle = 50 because An exterior angle of a triangle is equal to the sum of the opposite interior angles (pink angles).

Alternatively:

Green angle is 130 by sum of interior angles of a triangle.

Red angle is 50 by Adjacent angles

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What graph best represents the solution to the equality y - 2x > -8

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Answer:

<h2>In the attachment</h2>

Step-by-step explanation:

$y-2x>-8\qquad\text{add}\ 2x\ \text{to both sides}\\\\y>2x-8\\-------------------------\\\\-\ dotted\ line\\\leq,\ \geq-\ solid\ line\\,\ \geq-\ shading\ above\ the\ line\\-------------------------\\\\y=2x-8-\text{It's a linear function.}\\\text{We only need two points to draw a graph.}\\\text{Choice two arbitrary values of x, substitute to the equation,}\\\text{and calculate the values of y}.\\\\for\ x=4\to y=2(4)-8=8-8=0\to A(4,\ 0)\\for\ x=0\to y=2(0)-8=0-8=-8\to B(0,\ -8)$

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3 years ago

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3 years ago

John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart

Rina8888 [55]

<h2>Answer:</h2>

$Probability=\frac{1}{24}$

<h2>Step-by-step explanation:</h2>

As the question states,

John's brother has Galactosemia which states that his parents were both the carriers.

Therefore, the chances for the John to have the disease is = 2/3

Now,

Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

Now, one of those children married with a person.

So,

Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

Therefore,

<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

$Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}$

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)

<em><u>Hence, the probability for the first child to have Galactosemia is $\frac{1}{24}$ </u></em>

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3 years ago

Write with explain. ​

Write with explain.