I need help ASAP THANK YOU

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

Option 3

Step-by-step explanation:

We'll prove that their lengths are same by using distance formula.

Anestetic [448]3 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

To prove that BC ≅ AD, that is that BC = AD

You would need to prove the lengths are the same.

You might be interested in

Write the quadratic equation whose roots are −6 and 2 , and whose leading coefficient is 5 . (use the letter x to represent the

White raven [17]

<span>Here let the quadratic equation be ax^2 + bx + c We know that a=5 from the question. Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2). However, let's assume that a2=1 and b2=6, Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10 So, the equation is (5x-10)(x-6)</span>

3 0

3 years ago

The diagram shows triangles ABC and PQR. ABC is mapped to PQR by combining two single transformations

levacccp [35]

Try maybe just try hello reply back asap

7 0

2 years ago

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow

prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

$95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\$

$=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value $t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$